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Students can Download Maths Chapter 11 Conic Sections Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Question Bank Chapter 11 Conic Sections

Question 1.
Name the different conics.
Answer:
Circle, parabola, ellipse and hyperbola.

Circles

Question 2.
Define a circle and derive the equation of the circle whose centre and radius is given.
Answer:
Definition: A circle is a set of all points in a plane that are equidistant from a fixed point in a plane. OR A circle is the locus of a point which moves in a plane such that its distance from a fixed point in the plane is constant.
Note:

• Fixed point is called centre.
• The distance from the centre to a point on the circle is called radius.

Equation of a circle, given its centre and radius:
Let C(h,k) be the centre y and r, the radius of the circle. Let P(x,y) be any point on the circle. Then, by definition, |CPl = r
∴ By distance formula, we have
\(\sqrt{(x-h)^{2}+(y-k)^{2}}=r\)
\(\text { i.e., }(x-h)^{2}+(y-k)^{2}=r^{2}\)

Note: If the centre is at the origin ,then h = 0,k = 0. Therefore the equation of the circle with centre at the origin and radius r is x² + y² = r²

Question 3.
Find the equation of the circle with
(i) Centre (0,0) and radius r
(ii) Centre (0,2) and radius 2
(iii) Centre (-2,3) and radius 4
(iv) Centre (-3,2) and radius 4.
(v) Centre (1,1) and radius \(\sqrt{2}\)
(vi) \(\text { Centre }\left(\frac{1}{2}, \frac{1}{4}\right)\) and radius \(\frac{1}{12}\)
(viii) Centre (-a,-b) and radius \(\sqrt{a^{2}-b^{2}}\)
Answer:

Question 4.
Write the standard form (General form) of a circle.
Answer:
Standard form of a circle is x² + y² + 2 gx + 2 fy + c = 0
(∵ in the above problems, we observe that the equations of the circles are of the form
x² + y² + ( )x + ( )y + () = 0

Note:

• Centre = (-g, -f)
  \(\text { Radius }=\sqrt{g^{2}+f^{2}-c}\)
• The co-efficients of x² and y² are equal to unity.
• There is no term containing the product xy.
• If g² + f² -c > 0, then the circle is real circle.
• If g² + f² – c = 0, then the circle is a point circle.
• If g² + f² – c< 0, then the circle is imaginary circle.
• If the centre of the circle is on the x-axis, then equation of the circle is
  x² + y² + 2gx + c = 0 and conversely. (Here f = 0)
• If the centre of the circle is on the y-axis, then the equation of circle is
  x² + y² + 2fy + c = 0 and conversely. (Here g = 0)
• In the equation, x² + y² + 2gx + 2fy + c = 0, if c = 0, then the circle passes through the origin and circle equation is x² + y² + 2gx + 2fy = 0

Question 5.
Find the centre and radius of the circles.
(i) x² + y² + 2gx + 2fy + c = 0
(ii) x² + y² + 8x + 10y -8 = 0
(iii) x² + y² – 4x – 8y – 45 = 0
(iv) x² + y² – 8x – 10y -12 = 0
(v) (x + 5)² + (y – 3)² = 36
(vi) 2x² + 2y²-x = 0
Answer:

Question 6.
Find the equation of the circle passing through the points (2, -2) and (3, 4) and whose centre lies on the line x + y = 2
Answer:

Question 7.
Find the equation of the circle
(i) passing through the points (4, 1) and (6, 5) and whose centre lies on 4x + y = 16.
(ii) Passing through the points (2, 3) and (-1, 1) and centre is on the line x – 3y -11 = 0.
Answer:

Question 8.
Find the equation f the circle with radius 5 whose centre lies on x-axis and passing through the point (2,3).
Answer:

Question 9.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Answer:

Question 10.
Find the equation of the circle with centre (2,2) and passes through the point (4,5).
Answer:

Question 11.
Does the point (-2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?
Answer:

Parabola:

Question 1.
Define parabola.
Answer:
A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane. OR Parabola is the locus of a point which moves in a plane such that its distance from a fixed point is equal to its distance from a fixed line. The fixed point is called focus and fixed line is called directrix of the parabola.

Question 2.
Write the standard equations of parabola and their respective graphs
Answer:

Note: Parabolas are symmetrical about co-ordinate axes.

Question 3.
Derive the equation of a parabola in the form y² =4ax
Answer:
Let F be the focus and l be the directrix. Draw FM perpendicular to l. Take the  mid-point of FM as origin O and introduce the co­ordinate axes shown in the figure. Let the distance between the directrix and focus be 2a. (i.e., FM = 2a) Then |OF| = |OM| = a and F = (a,0) and equation of directrix is x = -a i.e., x + a =0.

Question 4.
Define latus rectum of a parabola and find the length of the latus rectum of the parabola y² = 4
Answer:
Definition: Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola.
Length of latus rectum of y² = 4ax:

By definition of parabola, AF = AC . But
FM = 2a = AC
∴  AF – 2a
Since the parabola is symmetric with respect to x-axis then
AF = FB = 2a
By definition, AB = latus rectum
∴ AB = AF + FB = 2a + 2a = 4a
∴  Length of latus rectum = 4a.

Note: Observations: (Useful for solving exercise) Vertex = (0, 0)

Equation	Axis	Focus	Directrix	LR
y2 = 4ax	x-axis	(a,0)	x + a = 0	4a
y2 =-4ax	x-axis	(-a, 0)	x-a = 0	4a
x2 = 4ay	y-axis	(0 ,a)	y + a = 0	4 a
x2 – 4ay	y-axis	(0 -a)	y – a = 0	4 a

Question 5.
Find the coordinates of the focus axis of the parabola, equation of the directrix and length of latus rectum.
(i) y² = 8x
(ii) y² =12x
(iii) y² = 10x
(iv) y² = – 8x
(v) x²  6y
(vi) x² = – 16y
(vii)  x² = -9y
(viii) 3y¹ = 12x
(ix)   3x² = -4y
(x) 2y² = -6x (viii, ix, x are not in exercise)
Answer:
Given (i) y² = 8x. This is of the form
y² = 4 ax
∴ 4a = 8 ⇒ = 21
We have, Focus: (a,0) = (2,0)
Axis: x-axis
Equation of directive: x + a = 0 i.e., x + 2=0
Length of latus rectum: 4a = 4(2) =8

(ii) Given: y² = 12x
Form: y² = 4ax
∴ 4a = 12 ⇒ a = 3
We have,
Focus : (a,0)=(3,0)
Axis : x – axis
Directrix: x + a=0 i.e., x + 3 = O
Latus rectum: 4a = 4(3) = 12

(iii) Try yourself.
\(F=\left(\frac{5}{2}, 0\right), x \text { -axis, } x+\frac{5}{2}=0,10\)

(iv) Given y² = – 8x
Form: y² = – 4ax
4a = 8 ⇒ a=2
We have,
Focus: ( – a,0) = ( – 2,0)
Axis: x-axis
Directrix : x – a = 0 ⇒ x + 2 = 0

Question 6.
Find the equation of the parabola that satisfies the give conditions.

(I)
(i) Vertex (0, 0) passing through (2, 3) and axis is along x-axis.
(ii) Vertex (0, 0) passing through (5, 2) and symmetrix with respect toy – axis.
(iii) Symmetric about y-axis and passing through the point (2, -3).
Answer:
I. (i) Given that
Focus = (2, 0), Directrix : x = – 2

Since (2, 0) lies on the positive then axis of the required parabola is x-axis!
∴ By data, required parabola is
y² – 4ax ………… (1)
We have, focus = (a,0) = (2,0) (given)
∴ a = 2
∴ (1) = y² = 4(2)x
⇒ y² = 8x  is the required parabola.

(ii) y2 = 24x (Try yourself)

(iii) Given, F= (0,- 3), l : y =3
Since F lies on the negative y-axis then axis of the parabola is y-axis.
By data, required parabola is
x² = – 4ay …………. (1)
But F = (0, – 3) = (0, – a)
∴ a = 3
(1) = x² = -4(3)y ⇒ x² = – 12y
is the required equation of the parabola.

(II)
(i) Given that, vertex = (0, 0) and focus = (0, 2)
Since focus lies on the positive y-axis then the required equation of the parabola is
x² = – 4ay ……………….(1)
Since, focus = (0,2) = (0,a) ∴ a = 2
∴ (1) ⇒ x² = – 4(2)y
⇒ x² = – 8y is required equation

(ii) Given that, vertex = (0, 0) and focus = (3,0)
Since focus lies on the positive x-axis, then the required equation of the parabola is
y² =4ax ………………. (1)
Since focus = (3, 0) = (a, 0) a = 3
(1) ⇒ y² = 4(3)x ⇒ y² = 12x

(iii) Given that, vertex = (0, 0) and focus = (-2, 0)
Since focus lies on the negative x-axis, then the required equation of the parabola is
y² = -4ax ………………. (1)

Since focus = (-2, 0) = (-a, 0) a- 2
(1) ⇒ y² = -4(2)x =5.1
y² = -8x

(III)
(i) Given that, Axis: x – axis
Vertex = (0, 0) and point = (2, 3)
Since (2, 3) lies in the first quadrant and axis is x-axis, then the required equation of the parabola is
y² =4ax ………………. (1)
Since (1) passes through the point (2, 3) then, we have

(ii) Given: V = (0, 0), Axis: y-axis, point = (5, 2) Since the point (5, 2) lies in the first quadrant and axis is y-axis then, the required equation of the parabola is
x² =4ay ………………. (1)
Since (1) passing through the point (5, 2) then, we have

(iii) Given: Axis: y-axis, point = (2,-3)
Since (2, -3) lies in the fourth quadrant and axis is y-axis then, the required equation of the parabola is x² = -4ay ………………. (1)
Since (1) passing through the point (2, -3) then, we have

Ellipse

Question 1.
Define ellipse.
Answer:
An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called foci.

Points to be noted:

• The mid point of the line segment joining the foci is called centre of the ellipse.
• The line segment through the foci is called the major axis.
• The line segment through the centre and perpendicular to the major axis is called minor axis.
• The ends points of major axis are called vertices of the ellipse.
• Length of major axis is 2a
• length of minor axis is 2b
• Distance between the foci is 2c
• Relationship between semi-major axis, semi­minor axis and distance of the focus from the centre of the ellipse is a² =b² + c² i.e., \(c=\sqrt{a^{2}-b^{2}}\)

Question 2.
Define eccentricity of an ellipse.
Answer:
The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse. Eccentricity is denoted by \(e.e=\frac{c}{a} \)(Here c <a ∴ e< 1)

Question 3.
Write standard equations of an ellipse and their respective graphs.
Answer:
Standard equations of an ellipse are  \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Question 4.
Derive the equation of an ellipse in the form
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:
Let F₁ and F₂ be the
foci and ‘O’ be the midpoint of F₁F₂
Let ‘O’ be the origin and line from ‘O’ <­through F₁ and F₂ be x-axis and through ‘O’ perpendicular to the x-axis be the y-axis.

Let F₁ =(c,0) and F₂ =(-c,0)
Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be ‘2a’. i.e., PF₁ + PF₁=2a ………………… (1)
By distance formula, we have

Question 5.
Define latus rectums of an ellipse and find the length of the latus rectum of the ellipse
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:
Definition: Latus rectum of an ellipse is a line segment perpendicular to Q the major axis through any of the foci and whose ends points lie on the ellipse. In the figure, AB and CD are latus rectum. Let the length of AF₁ be l . Then A = (c,l)

Question 6.
Find the coordinates of foci, the vertices, length of major and minor axes, the eccentricity and the latus rectum of the ellipse.

Answer:

Question 7.
Find the equation for the ellipse that satisfies the given conditions:
(i) Vertices (±5,0), foci (±4,0)
(ii) Vertices (0,±13), foci (0,±5)
(iii) Vertices (±13,0), foci (±5,0)
(iv) Ends of major axis (±3,0) ends of minor axis (0,±2)
(v) Ends of major axis \((0, \pm \sqrt{5})\), ends of minor axis (±1,0)
(vi) Length of major axis 26, foci (±5,0)
(vii) Length of major axis is 20, foci (0,±5)
(viii) Length of minor axis is 16, foci (0,±6)
(ix) Foci (±3,0), a = 4
(x) b = 3, c = 4 centre at the origin; foci on the x-axis.
(xi) Centre (0, 0), major axis on the y-axis and passes through the point (3,2) and (1,6)
(x) Major axis on the x-axis and passes through the points (4,3) and (6,2).
Answer:
(i) Given vertices Vertices (±5, 0), foci (±4,0)
We have, vertices = (±a,0) = (±5,0) a = 5
And foci = (±c, 0) (±4,0) c = 4
We have a² =b² + c²
i.e., b² =a² -c²
=(5)² -(4)² =25-16 = 9
i.e., b² = 9 ⇒ b = ±3
∴ a = 5 and b = 3
∴ Required equation of ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad( \text { foci lie on } x \text { -axis })\)
\(\text { i.e., } \sqrt{\frac{x^{2}}{25}+\frac{y^{2}}{9}=1}\)

(ii) Given Vertices = (0,±13), foci = (0,±5)
Since foci lie on y-axis, then required equation of the ellipse is \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) …………(1)

(iii) Given : Vertices = (±13,0), foci= (±5,0)
Since foci lie on the x-axis, then required equation of the ellipse is
\(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\)
Since, vertices = (±a, 0) = (±13,0) ∴ a =13

(iv) Given: Ends of major axis (±3,0) ends of minor axis (0,±2)
i.e., (± a,0) = (±3, 0) and (0, ±b) = (0,±2)
∴ a = 3 and b = 2
Required equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { i.e., } \frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)

(v) Given: Ends of major axis \((0, \pm \sqrt{5})\), ends of minor axis (±1,0)
Since ends of major axis lie on the y-axis, then required equation of the ellipse is,

(vi) Given: Length of major axis 26, foci (±5,0) Since, foci lie on the x-axis, t hen required equation of the
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
We have, length of major axis =2a = 26 (given)
and foci = (±c, 0) = (±5,0)
∴a=\(\frac{26}{2}=13\)
and foci = (±c, 0) = (±5,0)
∴ c = 5

(vi) Given: Length of minor axis is 16, foci (0,±6)
Since, foci lie on the y-axis, then required equation of the ellipse is

(viii) Given: Length of minor axis is 16, foci (0,±6)
Since, foci lie on the y-axis, then required equation of the ellipse is

(ix) Given: Foci (±3,0), a = 4
Since foci lie on the x-axis, then the required equation of the ellipse is

(x) Given: b = 3, c = 4 a = (0,0)
Since foci on the x-axis, then required equation of the ellipse is,

(xi) Given: Centre (0, 0), axis : y-axis
Since major axis is on the y-axis then, required equation of the ellipse is,
\(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\)
Since (1), passes through the point (3, 2) and (1,6) then, we

Hyperbola:

Question 1.
Define hyperbola.
Answer:
A hyperbola is the set of all points in a plane, the difference of whose distance from two fixed points in the plane is a constant.

Note: Observations

(i) The two fixed points F₁, F₂ are called foci of the hyperbola.
(ii) The mid-point of the line segment joining the foci is called centre of the hyperbola.
(iii) The line through foci is called transverse axis and the line through the centre and perpendicular to the transverse axis is called conjugate axis
(iv) The points at which the hyperbola intersects the transverse axis are called vertices of the hyperbola.
(v) Distance between foci is denoted by 2c, distance between vertices by 2a, and define the quantity b as \(b=\sqrt{c^{2}-a^{2}} \text { i.e., } \sqrt{b^{2}=c^{2}-a^{2}}\)
(vi) Length of transverse axis = 2a
Length of conjugate axis = 2b
(vii) Eccentricity = \(e=\frac{c}{a}(\text { Here } c \geq a)\)
∴ Eccentricity is never less than 1.

Standard equation of hyperbola and their graphs:

Question 2.
Derive the standard equation of the hyperbola in the form
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
Answer:

Let F₁ and F₂ be the foci and ‘O’ be the midpoint of line segment F₁F₂.
Let ‘O’ be the origin and the line through ‘O’, F₁ and F₂ be the x-axis. The line through O, perpendicular to the x-axis be the y-axis.
Let F₁= (c,0) and F₂ = (-c,0)
Let P = (x,y) be any point on the hyperbola such that the difference of the distances from P to the
F₁ and F₂ be 2a.
i.e., PF₂-PF₁=2a
Using the distance formula, we have

Length of latus rectum \(=\frac{2 b^{2}}{a}\)
Note: A hyperbola in which a = b is called an equilateral hyperbola.

Question 3.
Define latus rectum of hyperbola.
Answer:
Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola.

Question 4.
Find the coordinates of the foci, vertices, eccentricity, length of transverse axis and conjugate axis and length of latus rectum of the hyperbola.

Answer:

Question 5.
Find the equations of the hyperbola satisfying the given conditions.
(i) Vertices (±2,0, foci = (± 3,0)
(ii) Vertices = (0, ± 5), foci = (0, ± 8)
(iii) Vertices = (0, ± 3), foci =(0, ± 5)
(iv) Foci = (± 5,0) length of transverse axis = 8
(v) Foci = (0, ±13), length of conjugate axis = 24.
(vi) Foci = \((\pm 3 \sqrt{5}, 0) \), length of latus rectum = 12
(vii) Foci = (± 4,0) length of latus rectum -12
(viii) Foci =\((\pm 7,0), e=\frac{4}{3}\)
(ix) Foci = \((0, \pm \sqrt{10}) \)passing through (2,3)
(x) Foci = (0, ±3) vertices \(=\left(0, \pm \frac{\sqrt{11}}{2}\right)\)
(xi) Foci = (0, ± 12), length of latus rectum = 36.
Answer:
(i) Given, Vertices (±2,0, foci =(±3,0)
Since foci lie on the x-axis then the equation of the hyperbola is of the form
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \ldots(1)\)
We have,
vertices = (±a, 0) = (±2, 0) ∴ a =2
Also foci = (± c,0) = (±3,0)∴ c = 3
Since c² = a² +b²
∴ b² =c² – a² = (3)² – (2)² =9-4 = 5
∴ (1) \(\Rightarrow \frac{x^{2}}{4}-\frac{y^{2}}{5}=1\)

(ii) Given Vertices = (0, ± 5), foci = (0, ± 8)
Since foci lie on the y-axis then the required equation of the hyperbola is of the form

(iii) \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\) (Try yourself)

(iv) Given foci = (± 5, 0) T-axis = 8
Since, foci lie on the x-axis then the required equation of the hyperbola is of the form 2
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \) We have,
foci =(±c,0) = (±5,0) ∴ c = 5
Length of T-axis = 2a = 8 ∴ a = 4
Since b²=c²-a² =25-16 = 9

(v) Given foci = (0, ±13), c-axis length = 24.
Since foci lie on the y-axis then the equation of the hyperbola is of the form

(vi) Given foci \(=(\pm 3 \sqrt{5}, 0) \) length of LR = 8
Since foci lie on the x-axis then the required equation of the hyperbola is of the form

Miscellaneous examples

Question 1.
If a parabolic reflector is 20 cm in diameter and 5 cm deep find the focus.
Answer:
Given: AB = 20 cm, OC =5 cm

∴ AC – BC = 10 cm (curve is symmetric)
∴ A = (5,10), B = (5, – 10), C = (5, 0)
Let the equation of the parabola be
y² =4ax
Since A (5, 10) lies on (1), then we have,
(10)² = 4a(5)
∴100 = 20 a ∴ a =5
∴ Focus = (a, 0) = (5, 0) = c, the mid point of the diameter.
i.e., Focus of the reflector is at the mid point of the given diameter.

Question 2.
An arch is in the form of a parabola with its axis vertical. The arch is 10 m in high and 5 m wide at the base. How wide is it 2m from the vertex of the parabola?
Answer:
Given that AB = 5m, OC = 10m OD = 2m
Clearly, AC = CB = = \(\frac{5}{2} \)because the curve is symmetric about y-axis.

Question 3.
The cable of a uniformly loaded suspension bridge hands in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Answer:
Cable is in the form of an upward parabola, its equation is
x² = 4 ay ……………………. (1)

Question 4.
An arch is in the form of a semi ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.
Answer:
Equation of the arch be

Required height of the arch at a point Q, 1.5 m from one end A is 1.56 m.

Question 5.
A rod of length 12 cm moves with its ends always touching the co-ordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.
Answer:
Let AB = 12 cm be the rod at any position meet x-axis at A (a, 0) and y-axis at B (0, b).

Question 6.
Find the area of the triangle formed by the line joining the vertex of the parabola jc² = 12y to the ends of its latus rectum.
Answer:

Question 7.
A man running a race source notes that the stun of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.
Answer:
Required equation is of the form
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Question 8.
An equilater triangle is inscribed in the parabola y² = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Answer:
Given: ΔAOB is an equilateral triangle inscribed in the parabola y² = 4ax,
whose vertex is ‘O’

Question 9.
The focus of a parabolic mirror as shown in the figure is at a distance of 5 cm from its vertex. If the mirror is 45 cm deep, find the distance AB.
Answer:
From the given figure, we have equation of the parabolic mirror is
y² = 4ax ……………..(1)

Given: OF =5, F be the focus of (1)
Since F = (a, 0) = (5,0)  ∴  a = 5
And OC = 45 = Depth of the mirror.
A = (45,y) and B = (45,-y)
We have, x = 45, y = y and a = 5 then equation (1) gives,
y¹ =4 x 5 x 45 = 900 y = ± 30
AB = 2y = 2 x 30 = 60 cm

Question 10.
A rod AB, 15 cm rests in between two co-ordinate axes such that A lies on x-axis and B lies on y-axis. A point P (or, y) is a point on AB such t hat AP = 6 Show that the locus of P is an ellipse.
Answer:
Given: AB = 15, AP = 6 and PB = 9

Validating statements :

Validity of a statement (Truthfulness of a statement) can be established by the following methods.

• Direct method
• Contradiction method
• Contrapositive method
• Counter example method

Note: In mathematics, counter examples are used to disprove the statements.

Question 1.
Show that the statement p: “If x is a real number such that x³ + 4x = 0, then x is O” is true by
(i) Direct method,
(ii) Method of contradiction
(iii) Method of contrapositive.
Answer:
Given that, p: “If x is a real number such that x + 4x = 0, then x is O”.
(i) Direct method:
Let x³+4x = 0, x ∈ R
⇒ x(x² + 4) = 0, x ∈ R
⇒ x = 0 ∵ x² + 4 = 0
⇒ \(x=\pm \sqrt{-4} \notin R\)
⇒ p is a true statement.

(ii) Method of contradiction:
Let x ≠ O, x ∈ R
i.e., Let x be a nonzero real number
⇒ x² > 0 (y square of non-zero real number is always positive)
⇒ x² + 4 > 0 + 4
⇒ x²+ 4 > 4 x² + 4 ≠ 0
⇒ x(x² +4) ≠ 0
⇒ x³ + 4x = 0, which is a contradiction for x³ + 4x = 0 .
Hence x = 0 ⇒ p is true.

(iii) Method of contrapositive
∵ p: “If x is a real number such that x³ + 4x = 0, then x is O”.
Let q : “x is a real number and x³ + 4x = 0″ r:”x is 0″
∴  p is q ⇒ r
Its contrapositive is ~ r ⇒ ~ q
i.e., “I-x is non-zero real number, then x³ + 4x is also non-zero”.
Now, x ∈ R x 0
⇒ x² >0
⇒ x²+4>0 + 4
⇒ x² + 4 > 4
⇒ x² + 4 ≠ 0
⇒ x(x²+4)≠ 0
⇒ x³ + 4x ≠ 0
i.e., ~ r ⇒~ q is always true.
Hence q ⇒ r is true, i.e., p is true.

Question 2.
Show that the statement “For any real numbers a and b, a² = b² implies that a = b” is not true by giving counter example.
Answer:
Let a = 2 and b = – 2, then a and b are real numbers and a² = b²
∵ a² = (2)² = 4, b² = (-2)² = 4
a² = b² but a ≠ b (∵ 2 ≠ – 2)
Hence, a,b ∈ R and a² = b²
⇒ a = b is not true.

Question 3.
Show that the following statement is true by the method of contrapositive, p : If x is an integer and x² is even, then x is also even.
Answer:
Let q : x ∈ I and x² is even
r: x is even integer.
∴ Given p is q ⇒ r
Its contrapositive is ~ r  ⇒  ~ q.
Now, ~ r: x is not an even integer i.e., ~ r
⇒ x is an odd integer.
⇒x = 2n + 1 for n ∈ 1
⇒ x² = (2n +1)² = 4n² + 4n +1 = 2(2n² + 2n) +1
⇒ x² is an odd integer.
⇒  x² is not an even integer.
⇒ ~q
∴ ~ r ⇒ ~ q is true.
Hence q ⇒ r is true, i.e., p is true.

Question 4.
By giving counter example, show that the following statements are not true.
(i) If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.(ii) The equation x² – 1 = 0 does not have a root lying between 0 and 2.
Answer:
(i) Let PQR be triangle in which P = 60°, Q = 60°, R = 60°, then the triangle PQR is not an obtuse angled triangle. ∴  Given statement is not true.
(ii) Clearly x² -1 = 0 has x = 1 as one of the root lies between 0 and 2.
∴ Given statement is not true.

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Karnataka 1st PUC Maths Question Bank Chapter 12 Introduction to Three Dimensional Geometry

Question 1.
Define coordinate planes and rectangular coordinate system in space (or 3-D space).
Answer:
Let X’OX, Y’OY and Z’OZ be three mutually perpendicular lines having the same origin ‘O’. These lines are called X-axis, Y-axis and Z-axis respectively. The plane containing X’OX and Y’OY is called XOY plane (or XY-plane). i.e., the X-axis and Y-axis determine a plane, called XY-plane. the plane containing YOY’ and Z’OZ is called YOZ plane (or YZ-plane) and the plane containing X’OX and Z’OZ is called ZOX plane (or ZX – plane). These planes are called coordinate planes. This system of coordinate axes is called a rectangular coordinate system in space.

Definition: Coordinate planes:
Answer:
The three planes determined by the pair of axes are called (coordinate planes)

Note:

(i) The coordinates of a point P in 3-D is always written in the form of triplet like
(x, y, z), where x, y and z are the distances from the YZ, ZX and XY planes.

(ii) Coordinates of the origin O = (0, 0, 0)
(iii) Any point on the x-axis is of the form (x, 0, 0).
(iv) Any point on the y-axis is of the form (0, y, 0)
(v) Any point on the z-axis is of the form (0, 0, z)
(vi) Equations of x-axis are y = 0 and z = 0
(vii) Equations of y-axis are x = 0 and z = 0 (viii) Equations of z-axis are x = 0 and y = 0
(viii) Equation of XY-plane is z = 0 and any point in XY – plane will be (x, y, 0)
(ix) Equation of YZ – plane is x = 0 and any point in YZ – plane will be (0, y, z).
(x) Equation of ZX – plane is y = 0 and any point in the ZX – plane is (x, 0, z).

Question 2.
Define octants.
Answer:
The three coordinate planes divide the space into eight parts known as octants.

Question 3.
Write the table which shows the signs of the three coordinates in all the eight octants.
Answer:

Octant coordinate	I	II	III	IV	V	VI	VII	VII
X	+	–	–	+	+	–	–	+
y	+	+	–	–	+	+	–	–
z	+	+	+	+	–	–	–	–

Question 4.
Show that the distance between the points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\)
Answer:
Let P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) be the given points.

Let PQ be the line not parallel to z-axis. Draw PM and QN perpendiculars to XY-plane. Since XY-plane is two dimensional (2 – D), then the coordinates of M and N are respectively (x₁ , y₁) and (x₂ , y₂).

Section formula :

Question 5.
If A(x₁,y₁,z_l) and B(x₂,y₂,z₂) are two distinct points in space and a point P(x,y,z) divides AB in the ratio m : n internally then Prove that.
\(x=\frac{m x_{2}+n x_{1}}{m+n}, \quad y=\frac{m y_{2}+n y_{1}}{m+n}, \quad z=\frac{m z_{2}+n z_{1}}{m+n}\)
\(\text { i.e., } P=\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}, \frac{m z_{2}+n z_{1}}{m+n}\right)\)
Answer:
Draw AL, PN and BM perpendiculars to the xy-plane. Clearly the feet of the perpendiculars L, M, N are collinear and L = (x₁,y₁), M -(x₂,y₂), N = (x,y). Now, draw a line through P which is parallel to the line LM meets AL produced at C and BM at D. Clearly the triangle PAC and PDB are equiangular and hence similar.

Remark:
(i) It P divides AB externally in the ratio m : n, then the coordinates of the point P are

Observations:

• The point P divides AB internally in the ratio k : 1 iff k > 0 (i.e., k is + ve)
• The point P divides AB externally in the ratio |k| : 1 iff k < 0 (i.e., k is – ve)

Question 6.
Find the co-ordinates of the centroid of the triangle whose vertices are (x₁,y₁,z₁) (x₂,y₂,z₂) and (x₃,y₃,z₃)
Answer:
Note: Centroid is the point of intersection of the medians of the triangle and it is denoted by G. Centroid divides the median in the ratio 2:1.

Let A = (x₁,y₁,z₁) , B = (x₂,y₂,z₂) and C = (x₃, y₃, z₃) be the vertices of the triangle ABC. Let AD be the median drawn from A. Then D is the mid-point of BC.
∴ D=\(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}, \frac{z_{2}+z_{3}}{2}\right)\)
Let G be the centroid of the ΔABC. median AD in the ratio 2 : 1.
∴ The coordinates of G are,

Question 7.
Find the ratio in which the
(i) xy – plane
(ii) yz – plane
(iii) zx – plane divides the line segment formed by joining the points (x₁,y₁,z₁) and (x₂,y₂,z₂).
Answer:
We have, the coordinates of a point dividing the line joining the points A (x₁,y₁,z₁) and
B (x₂,y₂,z₂) in the ratio k : 1 are,
\(P=\left(\frac{k x_{2}+x_{1}}{k+1}, \frac{k y_{2}+y_{1}}{k+1}, \frac{k z_{2}+z_{1}}{k+1}\right)\)

(i) If xy – plane divides a line segment joining the given points, A, B, then z – coordinate of P is zero.
\(\text { i.e., } \frac{k z_{2}+z_{1}}{k+1}=0 \Rightarrow k z_{2}+z_{1}=0 \Rightarrow k=\frac{-z_{1}}{z_{2}}\)
∴ xy-plane divides the line segment AB in the ratio -z₁ : z₂
Similarly,
(ii) The yz-plane in the ratio -x₁: x₂
(iii) The zx-plane in the ratio -y_{1 :} y₂

Question 8.
A point is on the x-axis. What are its y and z – coordinates?
Answer:
Since a point is on the x-axis then it is of the form (x,0,0)
∴  y-coordinate = 0 and z-coordinate = 0.

Question 9.
A point is in the xz-plane. What can you say about its y-coordinate?
Answer:
Since a point is in the xz-plane, then it is of the form (x,0, z). So its y-coordinate is zero.

Question 10.
Name the octants in which the following points he. (1, 2, 3), (2, 4, 5), (-3, 1, 2), (-3, 1, -2), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4,2, 5), (-3,-1,6), (-2,-4,-7).
Answer:

Sl.No	Point	Octant
(i)	(1,2, 3)	first octant
(ii)	(2, 4, 5)	first octant
(iii)	(-3,1,2)	second octant
(iv)	(-3,1,-2)	sixth octant
(v)	(4,-2, 3)	fourth octant
(vi)	(4,-2,-5)	eighth octant
(vii)	(4, 2,-5)	fifth octant
(viii)	(-4, 2, -5)	sixth octant
(ix)	(-4, 2, 5)	second octant
(x)	(-3, -1,6)	third octant
(xi)	(-2, -4, -7)	seventh octant

Question 11.
Find the distance between the following pairs of points.
(i) (1,-3,4) and (-4,1,2)
(ii) (2, 3, 5) and (4,3,1)
(iii) (-3,7,2) and (2, 4,-1)
(iv) (-1,3, -4) and (1,-3,4)
(v) (2,-1,3) and (-2,1,3)
Answer:
(i) Let A (1, -3, 4) and B = (-4, 1, 2) be the given points.
Distance between the points A and B is

Question 12.
Show that the following points are collinear.
(i) (-2,3,5), (1,2,3) and (7, 0,-1)
(ii) (1,2,4), (3, 8,8), (7,20,16)
Answer:
(i) Let A (-2, 3, 5), B (1, 2, 3) and C (7, 0, -1) be the given points. We know that points are said to be collinear if they lie on a line.

\(\text { Thus } A B+B C=\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}=C A\)

Question 13.
Verify the following:
(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
(iii) (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are the vertices of a parallelogram.
Answer:
(i) Let A = (0,7,-10), B = (l,6,-6) and C = (4,9,-6) be vertices of ΔABC.


i..e., opposite sides of a quadrilateral ABCD are equal, therefore ABCD is a parallelogram.

Question 14.
Are the points A (3, 6, 9), B (10, 20, 30) C (25, -41, 5). The vertices of a right angled triangle.
Answer:
By distance formula,
AB² = (10 – 3)² + (20 – 6)² + (30 – 9)²
= 49 + 196 + 441 = 686
BC² = (25 -10)² + (- 41 – 20)² + (5 – 30)²
= 225 + 3721 + 625 = 4571
CA² = (3 – 25)² + (6 + 41)² + (9 – 5)²
= 484 + 2209 + 16 = 2709
We find that AB² + CA² ≠ BC²
(∵ 3395 ≠ 4571)
∴ ABC is not a right angled triangle.

Question 15.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3,2,-1).
Answer:
Let A (1, 2, 3) and B (3, 2, -1) be the given points.
Let P(x, y, z) be any point equidistant from A and B, then

⇒ x²-2x + 1+ y² – 4y + 4 + z² + 2z + 1
⇒ x²-6x + 9+y²-4y + 4 + z² + 2z + 1
⇒ 4x – 8z = 0
⇒ x – 2z = 0 is the required equation.

Question 16.
Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (-4,0,0) is equal to 10.
Answer:
Given points are A (4, 0, 0) and B (-4, 0, 0).
Let P(x, y, z) be any point such that
PA + PB = 10
⇒ PA = 10 – PB

Question 17.
Find the equation of set of points P such that PA² + PB² = 2k², where A and B are the points (3,4,5) and (-1,3, -7), respectively.
Answer:
Let P = (x,y,z), B = (3,4,5) and B = (-1,3-7).
Now, PB² = (x +1)² + (y – 3)² + (z + 7)² and PA²
= (x – 3)² + (y – 4)² + (z – 5)²
Given that PA² + PB² = 2k², then
(x-3)² + (y-4)² +(z-5)² + (x + 1)² +(y-3)²+(z + 7)² = 2k²
⇒ x² – 6*+ 9+ y² – 8y + 16 + z² – 10z + 22 + x² +2x +1 +
y² – 6y + 9 + z² +14z + 49 = 2k.
⇒ 2x² + 2y² + 2z² – 4x – 14y + 4z + 109 = 2k²
⇒ 2x² + 2y² + 2z² – Ax – 14y + 4z = 2k² -109

Question 18.
Find the coordinates of the point which divides the line segment joining the points
(-2, 3, 5) and (1, -4, 6) in the ratio
(i) 2 : 3 internally
(ii) 2:3 externally.
Answer:
The given points are A (-2, 3, 5) and B (1, -4, 6).

Question 19.
Find the coordinates of the point which divides the line segment joining the points
(1, -2, 3) and (3, 4, -5) in the ratio 2 : 3
(i) internally and
(ii) externally.
Answer:
Try yourself.
\(\text { (i) }\left(\frac{9}{5}, \frac{2}{5}, \frac{-1}{5}\right) \text { (ii) }(-3,-14,19)\)

Question 20.
Given that P (3, 2, -4), Q (5, 4, -6) and R (9, 8, -10) are collinear. Find the ratio in which Q divides PR.
Answer:
Let Q divides PR in the ratio k : 1, then Q is of the form,
\(Q=\left(\frac{k x_{2}+x_{1}}{k+1}, \frac{k y_{2}+y_{1}}{k+1}, \frac{k z_{2}+z_{1}}{k+1}\right)\)

Question 21.
Find the ratio in which the yz-plane divides the line segment formed by joining the points (-2,4,7) and (3,-5, 8)
Answer:
Let yz – plane divides the line segment joining A (-2, 4, 7) and B (3, -5, 8) at P(x,y,z) in the ratio k : 1. Then the coordinates of P are

Question 22.
Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the yz-plane.
Answer:
Let yz – plane divides the line segment joining (4, 8, 10) and (6, 10, -8) at P (x, y, z) in the ratio k : 1, then

∴ yz – plan divided AB externally in the ratio 2 : 3 \(\left(\ k=\frac{-2}{3}<0\right)\)

Question 23.
Using section formula, show that the points A(2,-3,4), B(-1,2,1) and \(C\left(0, \frac{1}{3}, 2\right)\)
Answer:
A(2,-3,4), B(-1,2,1) and \(C\left(0, \frac{1}{3}, 2\right)\) be the ratio k : 1 ,Then coordinates of the points P are,

Question 24.
Find the coordinates of the points which trisects the line segment joining the points
Answer:

Given points P (4, 2, -6) and Q (10, -16, 6).
Let A and B be the points of trisection of PQ, then A divides PQ in the ratio 1 : 2 and B divides PQ in the ratio 2:1.

Miscellaneous Examples

Question 1.
Show that the points A (1,2, 3), B (-1, -2, -1), C (2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
Answer:
Given points are A (1, 2, 3), B (-1, -2, -1), C (2, 3, 2) and D (4, 7, 6).
To show ABCD is a parallelogram we need to show opposite sides are equal.
By distance formula, we have

Since, AB = CD and BC = DA. ABCD is a parallelogram.
Now, it is required to prove that ABCD is not a rectangle. For this, we show that diagonals AC and BD are unequal.

Question 2.
Three vertices of a parallelogram ABCD are A (3, -1, 2), B (1, 2, -4) and C (-1, 1, 2). Find the co-ordinates of the fourths vertex.
Answer:
Let the fourth vertex be D(x₁ , y₁, z₁) we know that diagonals of a parallelogram bisect each other. So the mid-point of the diagonal AC is same as the mid-point of the diagonal BD.
∴ By mid-point formula, we have
\(\left(\frac{3+(-1)}{2}, \frac{-1+1}{2}+\frac{2+2}{2}\right)=\left(\frac{x_{1}+1}{2}, \frac{y_{1}+2}{2}, \frac{z_{1}-4}{2}\right)\)

Question 3.
Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6,0,0).
Answer:
Given (points) vertices A are A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

Let AD, BE and CF be the medians of the ΔABC. Then, D, E and F are the mid-points of the sides BC, CA and AB respectively.
∴ By mid-point formula, we have

Question 4.
If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6),
Q (-4, 3b, -10)and R (8, 14, 2c), then find the values of a, b and c.
Answer:
Given vertices are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).
Let G be the centroid of the ΔPQR then G = (0, 0, 0) (given).

Question 5.
The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, -5, 7) and (-1, 7, -6) respectively, find the coordinates of the point C.
Answer:
Let C = (x,y,z) and centroid G = (1,1,1). Then

Question 6.
Find the coordinates of a point on the y-axis which are at a distance of \(5 \sqrt{2} \)from the point P (3, -2,5).
Answer:
Given point is P (3, -2, 8)
Let Q be the point on the y-axis then Q be (0, y, 0).

∴ Required point on the y – axis is either (0,2,0) or (0,-6, 0)

Question 7.
A point R with x-coordinate 4 lies on the line segment joining the points P (2, -3, 4) and Q (8, 0, 10). Find the coordinates of the point R.
Answer:
Since R lies on the line PQ, then R divides PQ in some ratio say k : 1.
∴ By section formula R is of the form,

Question 8.
If A and B be the points (3, 4, 5) and (-1, 3, -7) respectively, find the equation of the set of points P such that PA² +PB² =k² where k is constant.
Answer:
Given points are A (3, 4, 5) and B (-1, 3, -7).
Let P = (x,y,z)
Given: PA² + PB² = k²
⇒ (x-3)² + (y-4)² + (z-5)² +(x + 1)² + (y – 3)² + (z + 7)² =k²
⇒ 2x¹ – 6x + 9 + y² -8y + 16 + z² -10z + 25 + x² + 2x +
1+y² – 6y + 9 + z² +14z + 49 = k²
⇒ 2x² + 2y² + 2z² – 4x – 14y + 4z +109 = k²
∴ Required equations of the set of points P is,
2x² + 2y² +2z² – 4x – 14y + 4z + 109 – k² =0

Question 9.
Find the equation of the set of points P such that PA = PB, where A (3, 4, -5) and
B (-2,1,4).
Answer:
Try yourself. (10x – 6y – 18z – 29 = 0)

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