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Class 11 (FYJC / HSC)Mathematics & Statistics2026-27 Syllabus

Chapter 7 Conic Sections Ex 7.2 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 7 Conic Sections Ex 7.2. Step-by-step solved exercises, numerical problems, and digest answers.

18 Solved Questions• 25 Diagrams• 4160 words

Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.2

Question 1 Maharashtra Board Solution
Find the (i) lengths of the principal axes (ii) co-ordinates of the foci (iii) equations of directrices (iv) length of the latus rectum (v) distance between foci (vi) distance between directrices of the ellipse: (a) (b) 3x2 + 4y2 = 12 (c) 2x2 + 6y2 = 6 (d) 3x2 + 4y2 = 1
Solution & Step-by-Step Answer:
(a) Given equation of the ellipse is Comparing this equation with , we get a2 = 25 and b2 = 9 a = 5 and b = 3 Since a > b, X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(3) = 6
Lengths of the principal axes are 10 and 6.

(ii) We know that e =
=
=
=
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(5(), 0) and S'(-5(), 0)
i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±
=
=

(iv) Length of latus rectum =

(v) Distance between foci = 2ae
= 2(5)()
= 8

(vi) Distance between directrices =
=
=

(b) Given equation of the ellipse is 3x2+ 4y2= 12

Comparing this equation with , we get
a2= 4 and b2= 3
a = 2 and b = √3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(2) = 4
Length of minor axis = 2b = 2√3
Lengths of the principal axes are 4 and 2√3.

(ii) We know that e =
=
=
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(2(), 0) and S'(-2(), 0)
i.e., S(1, 0) and S'(-1, 0)

(iii) Equations of the directrices are x = ±
=
= ±4

(iv) Length of latus rectum =

(v) Distance between foci = 2ae = 2(2)() = 2

(vi) Distance between directrices =
=
= 8

(c) Given equation of the ellipse is 2x2+ 6y2= 6

Comparing this equation with , we get
a2= 3 and b2= 1
a = √3 and b = 1
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2√3
Length of minor axis = 2b = 2(1) = 2
Lengths of the principal axes are 2√3 and 2.

(ii) We know that e =
=
=
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(√3(), o) and S'(-√3(), 0)
i.e., S(√2, 0) and S'(-√2, 0)

(iii) Equations of the directrices are x = ±,
=
=

(iv) Length of latus rectum =

(v) Distance between foci = 2ae
=
= 2√2

(vi) Distance between directrices =
=
=
= 3√2

(d) Given equation of the ellipse is 3x2+ 4y = 1.

Comparing this equation with , we get
a2= and b2=
a = and b =
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2() =
Length of minor axis = 2b = 2() = 1
Lengths of the principal axes are and 1.

(ii) We know that e =
e =
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S and S’
i.e., S(, 0) and S'(-, 0)

(iii) Equations of the directrices are x = ±,
=
=

(iv) Length of latus rectum =
=
=

(v) Distance between foci = 2ae
=
=

(vi) Distance between directrices =
=
=

Question 2 Maharashtra Board Solution
Find the equation of the ellipse in standard form if (i) eccentricity = and distance between its foci = 6. (ii) the length of the major axis is 10 and the distance between foci is 8. (iii) distance between directrices is 18 and eccentricity is . (iv) minor axis is 16 and eccentricity is . (v) the distance between foci is 6 and the distance between directrices is . (vi) the latus rectum has length 6 and foci are (±2, 0). (vii) passing through the points (-3, 1) and (2, -2). (viii) the distance between its directrices is 10 and which passes through (-√5, 2). (ix) eccentricity is and passes through (2, ).
Solution & Step-by-Step Answer:
(i) Let the required equation of ellipse be , where a > b. Given, eccentricity (e) = Distance between foci = 2ae Given, distance between foci = 6 2ae = 6 The required equation of ellipse is .

(ii) Let the required equation of ellipse be , where a > b.
Length of major axis = 2a
Given, length of major axis = 10
2a = 10
a = 5
a2= 25
Distance between foci = 2ae
Given, distance between foci = 8
2ae = 8
2(5)e = 8

The required equation of ellipse is .

(iii) Let the required equation of ellipse be , where a > b.
Given, eccentricity (e) =
Distance between directrices =
Given, distance between directrices = 18

The required equation of ellipse is

(iv) Let the required equation of ellipse be , where a > b.
Length of minor axis = 2b
Given, length of minor axis = 16
2b = 16
b = 8
b2= 64
Given, eccentricity (e) =
Now, b2= a2(1 – e2)

The required equation of ellipse is .

(v) Let the required equation of ellipse be , where a > b.
Distance between foci = 2ae
Given, distance between foci = 6
2ae = 6
ae = 3
a = …….(i)
Distance between directrices =
Given, distance between directrices =

The required equation of ellipse is .

(vi) Given, the length of the latus rectum is 6, and co-ordinates of foci are (±2, 0).
The foci of the ellipse are on the X-axis.
Let the required equation of ellipse be , where a > b.
Length of latus rectum =
= 6
b2= 3a …..(i)
Co-ordinates of foci are (±ae, 0)
ae = 2
a2e2= 4 …..(ii)
Now, b2= a2(1 – e2)
b2= a2– a2e2
3a = a2– 4 …..[From (i) and (ii)]
a2– 3a – 4 = 0
a2– 4a + a – 4 = 0
a(a – 4) + 1(a – 4) = 0
(a – 4) (a + 1) = 0
a – 4 = 0 or a + 1 = 0
a = 4 or a = -1
Since a = -1 is not possible,
a = 4
a2= 16
Substituting a = 4 in (i), we get
b2= 3(4) = 12
The required equation of ellipse is .

(vii) Let the required equation of ellipse be , where a > b.
The ellipse passes through the points (-3, 1) and (2, -2).
Substituting x = -3 and y = 1 in equation of ellipse, we get

Equations (i) and (ii) become
9A + B = 1 …..(iii)
4A + 4B = 1 …..(iv)
Multiplying (iii) by 4, we get
36A + 4B = 4 …..(v)
Subtracting (iv) from (v), we get
32A = 3
A =
Substituting A = in (iv), we get
4() + 4B = 1
+ 4B = 1
4B = 1 –
4B =
B =

(viii) Let the required equation of ellipse be , where a > b.
Distance between directrices =
Given, distance between directrices = 10
= 10
a = 5e …..(i)
The ellipse passes through (-√5, 2).
Substituting x = -√5 and y = 2 in equation of ellipse, we get

b2= 15()
b2= 6
The required equation of ellipse is .

(ix) Let the required equation of ellipse be , where a > b.
Given, eccentricity (e) =
The ellipse passes through (2, ).
Substituting x = 2 and y = in equation of ellipse, we get

The required equation of ellipse is .

Question 3 Maharashtra Board Solution
Find the eccentricity of an ellipse, if the length of its latus rectum is one-third of its minor axis.
Solution & Step-by-Step Answer:
Let the equation of ellipse be , where a > b. Length of latus rectum = Length of minor axis = 2b According to the given condition, Length of latus rectum = (Minor axis)

Question 4 Maharashtra Board Solution
Find the eccentricity of an ellipse, if the distance between its directrices is three times the distance between its foci.
Solution & Step-by-Step Answer:
Let the required equation of ellipse be , where a > b. Distance between directrices = Distance between foci = 2ae According to the given condition, distance between directrices = 3(distance between foci) = 3(2ae) = 3e = e2 e = ……[∵ 0 < e < 1] Eccentricity of the ellipse is
Question 5 Maharashtra Board Solution
Show that the product of the lengths of the perpendicular segments drawn from the foci to any tangent line to the ellipse is equal to 16.
Solution & Step-by-Step Answer:
Given equation of the ellipse is . Comparing this equation with , we get a2 = 25, b2 = 16 a = 5, b = 4

Question 6 Maharashtra Board Solution
A tangent having slope the ellipse 3x2 + 4y2 = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.
Solution & Step-by-Step Answer:
Given equation of the ellipse is 3x2 + 4y2 = 12. Comparing this equation with , we get a2 = 4, b2 = 3 Equations of tangents to the ellipse having slope m are y = mx ± Here, m = Equations of the tangents are y = 2y = -x ± 4 x + 2y ± 4 = 0 Consider the tangent x + 2y – 4 = 0 Let this tangent intersect the X-axis at A(x1, 0) and Y-axis at B(0, y1). x1 + 0 – 4 = 0 and 0 + 2y1 – 4 = 0 x1 = 4 and y1 = 2 A = (4, 0) and B = (0, 2) l(OA) = 4 and l(OB) = 2 Area of ∆AOB = × l(OA) × l(OB) = × 4 × 2 = 4 sq.units

Question 7 Maharashtra Board Solution
Show that the line x – y = 5 is a tangent to the ellipse 9x2 + 16y2 = 144. Find the point of contact.
Solution & Step-by-Step Answer:
Given equation of the ellipse is 9x2 + 16y2 = 144 Comparing this equation with , we get a2 = 16 and b2 = 9 Given equation of line is x – y = 5, i.e., y = x – 5 c2 = a2 m2 + b2 Comparing this equation with y = mx + c, we get m = 1 and c = -5 For the line y = mx + c to be a tangent to the ellipse 1, we must have c2 = a2 m2 + b2 c2 = (-5)2 = 25 a2 m2 + b2 = 16(1)2 + 9 = 16 + 9 = 25 = c2 The given line is a tangent to the given ellipse and point of contact = = =
Question 8 Maharashtra Board Solution
Show that the line 8y + x = 17 touches the ellipse x2 + 4y2 = 17. Find the point of contact.
Solution & Step-by-Step Answer:
Given equation of the ellipse is x2 + 4y2 = 17. Comparing this equation with , we get a2 = 17 and b2 = Given equation of line is 8y + x = 17, y = Comparing this equation with y = mx + c, we get m = and c = For the line y = mx + c to be a tangent to the ellipse 1, we must have

Question 9 Maharashtra Board Solution
Determine whether the line x + 3y√2 = 9 is a tangent to the ellipse . If so, find the co-ordinates of the point of contact.
Solution & Step-by-Step Answer:
Given equation of the ellipse is Comparing this equation with , we get a2 = 9 and b2 = 4 Given equation of line is x + 3y√2 = 9, i.e., y = Comparing this equation with y = mx + c, we get m = and c = For the line y = mx + c to be a tangent to the ellipse 1, we must have

Question 10 Maharashtra Board Solution
Find k, if the line 3x + 4y + k = 0 touches 9x2 + 16y2 = 144.
Solution & Step-by-Step Answer:
Given equation of the ellipse is 9x2 + 16y2 = 144. Comparing this equation with , we get a2 = 16 and b2 = 9 Given equation of line is 3x + 4y + k = 0, i.e., y = Comparing this equation with y = mx + c, we get m = and c = For the line y = mx + c to be a tangent to the ellipse 1, we must have c2 = a2 m2 + b2 = 9 + 9 = 18 k2 = 288 k = ±12√2
Question 11 Maharashtra Board Solution
Find the equations of the tangents to the ellipse: (i) passing through the point (2, -2). (ii) 4x2 + 7y2 = 28 from the point (3, -2). (iii) 2x2 + y2 = 6 from the point (2, 1). (iv) x2 + 4y2 = 9 which are parallel to the line 2x + 3y – 5 = 0. (v) which are parallel to the line x + y + 1 = 0. (vi) 5x2 + 9y2 = 45 which are ⊥ to the line 3x + 2y + 1 = 0. (vii) x2 + 4y2 = 20 which are ⊥ to the line 4x + 3y = 7.
Solution & Step-by-Step Answer:
(i) Given equation of the ellipse is . Comparing this equation with , we get a2 = 5 and b2 = 4 Equations of tangents to the ellipse having slope m are y = mx ± Since (2, -2) lies on both the tangents, -2 = 2m ± -2 – 2m = ± Squaring both the sides, we get 4m2 + 8m + 4 = 5m2 + 4 m2 – 8m = 0 m(m – 8) = 0 m = 0 or m = 8 These are the slopes of the required tangents. By slope point form y – y1 = m(x – x1), the equations of the tangents are y + 2 = 0(x – 2) and y + 2 = 8(x – 2) y + 2 = 0 and y + 2 = 8x – 16 y + 2 = 0 and 8x – y – 18 = 0

(ii) Given equation of the ellipse is 4x2+ 7y2= 28.

Comparing this equation with , we get
a2= 7 and b2= 4
Equations of tangents to the ellipse having slope m are
y = mx ±
Since (3, -2) lies on both the tangents,
-2 = 3m ±
-2 – 3m = ±
Squaring both the sides, we get
9m2+ 12m + 4 = 7m2+ 4
2m2+ 12m = 0
2m(m + 6) = 0
m = 0 or m = -6
These are the slopes of the required tangents.
By slope point form y – y1= m(x – x1),
the equations of the tangents are
y + 2 = 0(x – 3) and y + 2 = -6(x – 3)
y + 2 = 0 and y + 2 = -6x + 18
y + 2 = 0 and 6x + y – 16 = 0

(iii) Given equation of the ellipse is 2x2+ y2= 6.

Comparing this equation with , we get
a2= 3 and b2= 6
Equations of tangents to the ellipse having slope m are
y = mx ±
Since (2, 1) lies on both the tangents,
1 = 2m ±
1 – 2m = ±
Squaring both the sides, we get
1 – 4m + 4m2= 3m2+ 6
m2– 4m – 5 = 0
(m – 5)(m + 1) = 0
m = 5 or m = -1
These are the slopes of the required tangents.
By slope point form y – y1= m(x – x1),
the equations of the tangents are
y – 1 = 5(x – 2) and y – 1 = -1(x – 2)
y – 1 = 5x – 10 and y – 1 = -x + 2
5x – y – 9 = 0 and x + y – 3 = 0

(iv) Given equation of the ellipse is x2+ 4y2= 9.

Comparing this equation with , we get
a2= 9 and b2=
Slope of the line 2x + 3y – 5 = 0 is .
Since the given line is parallel to the required tangents, slope of the required tangents is
m =
Equations of tangents to the ellipse having slope m are

(v) Given equation of the ellipse is .
Comparing this equation with , we get
a2= 25 and b2= 4
Slope of the given line x + y + 1 = 0 is -1.
Since the given line is parallel to the required tangents,
the slope of the required tangents is m = -1.
Equations of tangents to the ellipse

(vi) Given equation of the ellipse is 5x2+ 9y2= 45.

Comparing this equation with , we get
a2= 9 and b2= 5
Slope of the given line 3x + 2y + 1 = 0 is
Since the given line is perpendicular to the required tangents, slope of the required tangents is
m =
Equations of tangents to the ellipse

(vii) Given equation of the ellipse is x2+ 4y2= 20.

Comparing this equation with , we get
a2= 20 and b2= 5
Slope of the given line 4x + 3y = 7 is .
Since the given line is perpendicular to the required tangents,
slope of the required tangents is m = .
Equations of tangents to the ellipse having slope m are
y = mx ±
y =

Question 12 Maharashtra Board Solution
Find the equation of the locus of a point, the tangents from which to the ellipse 3x2 + 5y2 = 15 are at right angles.
Solution & Step-by-Step Answer:
Given equation of the ellipse is 3x2 + 5y2 = 15. Comparing this equation with , we get a2 = 5 and b2 = 3 Equations of tangents to the ellipse having slope m are y = mx ± y = mx ± y – mx =± Squaring both the sides, we get y2 – 2mxy + m2x2 = 5m2 + 3 (x2 – 5) m2 – 2xym + (y2 – 3) = 0 The roots m1 and m2 of this quadratic equation are the slopes of the tangents. m1m2 = Since the tangents are at right angles, m1m2 = -1 y2 – 3 = -x2 + 5 x2 + y2 = 8, which is the required equation of the locus.

Alternate method:
The locus of the point of intersection of perpendicular tangents is the director circle of an ellipse.
The equation of the director circle of an ellipse is x2+ y2= a2+ b2
Here, a2= 5 and b2= 3
x2+ y2= 5 + 3
x2+ y2= 8, which is the required equation of the locus.

Question 13 Maharashtra Board Solution
Tangents are drawn through a point P to the ellipse 4x2 + 5y2 = 20 having inclinations θ1 and θ2 such that tan θ1 + tan θ2 = 2. Find the equation of the locus of P.
Solution & Step-by-Step Answer:
Given equation of the ellipse is 4x2 + 5y2 = 20. Comparing this equation with , we get a2 = 5 and b2 = 4 Since inclinations of tangents are θ1 and θ2, m1 = tan θ1 and m2 = tan θ2 Equation of tangents to the ellipse having slope m are y = mx ± y = mx ± y – mx = ± Squaring both the sides, we get y2 – 2mxy + m2x2 = 5m2 + 4 (x2 – 5)m2 – 2xym + (y2 – 4) = 0 The roots m1 and m2 of this quadratic equation are the slopes of the tangents. m1 + m2 = Given, tan θ1 + tan θ2 = 2 m1 + m2 = 2 = 2 xy = x2 – 5 x2 – xy – 5 = 0, which is the required equation of the locus of P.
Question 14 Maharashtra Board Solution
Show that the locus of the point of intersection of tangents at two points on an ellipse, whose eccentric angles differ by a constant, is an ellipse.
Solution & Step-by-Step Answer:
Let P(θ1) and Q(θ2) be any two points on the given ellipse such that θ1 – θ2 = k, where k is a constant. The equation of the tangent at point P(θ1) is ……(i) The equation of the tangent at point Q(θ2) is ……(ii) Multiplying equation (i) by cos θ2 and equation (ii) by cos θ1 and subtracting, we get (sin θ1 cos θ2 – sin θ2 cos θ1) = cos θ2 – cos θ1

Question 15 Maharashtra Board Solution
P and Q are two points on the ellipse with eccentric angles θ1 and θ2. Find the equation of the locus of the point of intersection of the tangents at P and Q if θ1 + θ2 = .
Solution & Step-by-Step Answer:
Given equation of the ellipse is . θ1 and θ2 are the eccentric angles of a tangent. Equation of tangent at point P is ……(i) Equation of tangent at point Q is ………(ii) θ1 + θ2 = …..[Given] i.e., points P and Q coincide, which is not possible, as P and Q are two different points. cos θ1 – sin θ1 ≠ 0 Dividing equation (iii) by (cos θ1 – sin θ1), we get bx1 – ay1 = 0 bx – ay = 0, which is the required equation of locus of point M.

Question 16 Maharashtra Board Solution
The eccentric angles of two points P and Q of the ellipse 4x2 + y2 = 4 differ by . Show that the locus of the point of intersection of the tangents at P and Q is the ellipse 4x2 + y2 = 16.
Solution & Step-by-Step Answer:
Given equation of the ellipse is 4x2 + y2 = 4. Let P(θ1) and Q(θ2) be any two points on the given ellipse such that θ1 – θ2 = The equation of the tangent at point P(θ1) is ……(i) The equation of the tangent at point Q(θ2) is Multiplying equation (i) by cos θ2 and equation (ii) by cos θ1 and subtracting, we get

Question 17 Maharashtra Board Solution
Find the equations of the tangents to the ellipse , making equal intercepts on co-ordinate axes.
Solution & Step-by-Step Answer:
Given equation of the ellipse is Comparing this equation with , we get a2 = 16 and b2 = 9 Since the tangents make equal intercepts on the co-ordinate axes, m = -1. Equations of tangents to the ellipse having slope m are y = mx ± y = -x ± y = -x ± x + y = ±5
Question 18 Maharashtra Board Solution
A tangent having slope to the ellipse 3x2 + 4y2 = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.
Solution & Step-by-Step Answer:
The equation of the ellipse is 3x2 + 4y2 = 12 Comparing with , we get a2 = 4, b2 = 3 The equation of tangent with slope m is It meets X axis at A ∴ for A, put y = 0 in equation (1), we get, x = ±4 ∴ A = (±4, 0) Similarly, B = (0, ±2) ∴ OA = 4, OB = 2 ∴ Area of ΔOAB = × OA × OB = × 4 × 2 = 4 sq. units

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