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

Chapter 7 Conic Sections Ex 7.3 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 7 Conic Sections Ex 7.3. Step-by-step solved exercises, numerical problems, and digest answers.

10 Solved Questions• 24 Diagrams• 2337 words

Maharashtra State Board 11th Maths Solutions Chapter 7 Conic Sections Ex 7.3

Question 1 Maharashtra Board Solution
Find the length of the transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices, and the length of the latus rectum of the hyperbolae. (i) (ii) (iii) 16x2 – 9y2 = 144 (iv) 21x2 – 4y2 = 84 (v) 3x2 – y2 = 4 (vi) x2 – y2 = 16 (vii) (viii) (ix) (x) x = 2 sec θ, y = 2√3 tan θ
Solution & Step-by-Step Answer:
(i) Given equation of the hyperbola is Comparing this equation with , we get a2 = 25 and b2 = 16 ⇒ a = 5 and b = 4 Length of transverse axis = 2a = 2(5) = 10 Length of conjugate axis = 2b = 2(4) = 8 We know that

(ii) Given equation of the hyperbola is

Comparing this equation with , we get
b2= 16 and a2= 25
⇒ b = 4 and a = 5
Length of transverse axis = 2b = 2(4) = 8
Length of conjugate axis = 2a = 2(5) = 10
Co-ordinates of vertices are B(0, b) and B’ (0, -b)
i.e., B(0, 4) and B'(0, -4)
We know that

(iii) Given equation of the hyperbola is 16x2– 9y2= 144.

Comparing this equation with , we get
a2= 9 and b2= 16
⇒ a = 3 and b = 4
Length of transverse axis = 2a = 2(3) = 6
Length of conjugate axis = 2b = 2(4) = 8
We know that
e =
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(3(), 0) and S'(-3(), 0)
i.e., S(5, 0) and S'(-5, 0)
Equations of the directrices are x = ±
=
=
Length of latus rectum =

(iv) Given equation of the hyperbola is 21x2– 4y2= 84.

Comparing this equation with , we get
a2= 4 and b2= 21
⇒ a = 2 and b = √21
Length of transverse axis = 2a = 2(2) = 4
Length of conjugate axis = 2b = 2√21
We know that

(v) Given equation of the hyperbola is 3x2– y2= 4.

Comparing this equation with , we get
a2= and b2= 4

(vi) Given equation of the hyperbola is x2– y2= 16.

Comparing this equation with , we get
a2= 16 and b2= 16
⇒ a = 4 and b = 4
Length of transverse axis = 2a = 2(4) = 8
Length of conjugate axis = 2b = 2(4) = 8
We know that
e =
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S (4√2, 0) and S’ (-4√2, 0)
Equations of the directrices are x = ±
⇒ x =
⇒ x = ± 2√2
Length of latus rectum = = 8

(vii) Given equation of the hyperbola is .
Comparing this equation with , we get
b2= 25 and a2= 9
⇒ b = 5 and a = 3
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(3) = 6
Co-ordinates of vertices are B(0, b) and B’ (0, -b),
i.e., B(0, 5) and B’ (0, -5)
We know that

(viii) Given equation of the hyperbola is .
Comparing this equation with , we get
b2= 25 and a2= 144
⇒ b = 5 and a = 12
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(12) = 24
Co-ordinates of vertices are B(0, b) and B’ (0, -b),
i.e., B(0, 5) and B’ (0, -5)
We know that

(ix) Given equation of the hyperbola is
Comparing this equation with , we get
a2= 100 and b2= 25
⇒ a = 10 and b = 5
Length of transverse axis = 2a = 2(10) = 20
Length of conjugate axis = 2b = 2(5) = 10
We know that

(x) Given equation of the hyperbola is x = 2 sec θ, y = 2√3 tan θ.
Since sec2θ – tan2θ = 1,


Comparing this equation with , we get
a2= 4 and b2= 12
⇒ a = 2 and b = 2√3
Length of transverse axis = 2a = 2(2) = 4
Length of conjugate axis = 2b = 2(2√3) = 4√3
We know that
e = = = 2
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(2(2), 0) and S'(-2(2), 0),
i.e., S(4, 0) and S'(-4, 0)
Equations of the directrices are x = ±.
⇒ x = ±
⇒ x = ±1
Length of latus rectum = = 12

Question 2 Maharashtra Board Solution
Find the equation of the hyperbola with centre at the origin, length of the conjugate axis as 10, and one of the foci as (-7, 0).
Solution & Step-by-Step Answer:
Given, one of the foci of the hyperbola is (-7, 0). Since this focus lies on the X-axis, it is a standard hyperbola. Let the required equation of hyperbola be Length of conjugate axis = 2b Given, length of conjugate axis = 10 ⇒ 2b = 10 ⇒ b = 5 ⇒ b2 = 25 Co-ordinates of focus are (-ae, 0) ae = 7 ⇒ a2e2 = 49 Now, b2 = a2(e2 – 1) ⇒ 25 = 49 – a2 ⇒ a2 = 49 – 25 = 24 The required equation of hyperbola is
Question 3 Maharashtra Board Solution
Find the eccentricity of the hyperbola, which is conjugate to the hyperbola x2 – 3y2 = 3
Solution & Step-by-Step Answer:
Given, equation of hyperbola is x2 – 3y2 = 3. Equation of the hyperbola conjugate to the above hyperbola is Comparing this equation with , we get b2 = 1 and a2 = 3 Now, a2 = b2(e2 – 1) ⇒ 3 = 1(e2 – 1) ⇒ 3 = e – 1 ⇒ e2 = 4 ⇒ e = 2 …..[∵ e > 1]
Question 4 Maharashtra Board Solution
If e and e’ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that .
Solution & Step-by-Step Answer:
Let e be the eccentricity of a hyperbola

Question 5 Maharashtra Board Solution
Find the equation of the hyperbola referred to its principal axes: (i) whose distance between foci is 10 and eccentricity is (ii) whose distance between foci is 10 and length of the conjugate axis is 6. (iii) whose distance between directrices is and eccentricity is . (iv) whose length of conjugate axis = 12 and passing through (1, -2). (v) which passes through the points (6, 9) and (3, 0). (vi) whose vertices are (±7, 0) and endpoints of the conjugate axis are (0, ±3). (vii) whose foci are at (±2, 0) and eccentricity is . (viii) whose lengths of transverse and conjugate axes are 6 and 9 respectively. (ix) whose length of transverse axis is 8 and distance between foci is 10.
Solution & Step-by-Step Answer:
(i) Let the required equation of hyperbola be Given, eccentricity (e) = Distance between foci = 2ae Given, distance between foci = 10 ⇒ 2ae = 10 ⇒ ae = 5 ⇒ a() = 5 ⇒ a = 2 ⇒ a2 = 4

(ii) Let the required equation of hyperbola be
Length of conjugate axis = 2b
Given, length of conjugate axis = 6
⇒ 2b = 6
⇒ b = 3
⇒ b2= 9
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a2e2= 25
Now, b2= a2(e2– 1)
⇒ b2= a2e2– a2
⇒ 9 = 25 – a2
⇒ a2= 25 – 9
⇒ a2= 16
The required equation of hyperbola is

(iii) Let the required equation of hyperbola be
Given, eccentricity (e) =
Distance between directrices =
Given, distance between directrices =

(iv) Let the required equation of hyperbola be
……(i)
Length of conjugate axis = 2b
Given, length of conjugate axis = 12
⇒ 2b = 12
⇒ b = 6 …..(ii)
⇒ b2= 36
The hyperbola passes through (1, -2)
Substituting x = 1 and y = -2 in (i), we get

(v) Let the required equation of hyperbola be
……(i)
The hyperbola passes through the points (6, 9) and (3, 0).

(vi) Let the required equation of hyperbola be

Co-ordinates of vertices are (±a, 0).
Given that, co-ordinates of vertices are (±7, 0)
∴ a = 7
Endpoints of the conjugate axis are (0, b) and (0, -b).
Given, the endpoints of the conjugate axis are (0, ±3).
∴ b = 3
The required equation of hyperbola is
i.e.,

(vii) Let the required equation of hyperbola be
……(i)
Given, eccentricity (e) =
Co-ordinates of foci are (±ae, 0).
Given co-ordinates of foci are (±2, 0)
ae = 2
⇒ a() = 2
⇒ a =
⇒ a2=

(viii) Let the required equation of hyperbola be

Length of transverse axis = 2a
Given, length of transverse axis = 6
⇒ 2a = 6
⇒ a = 3
⇒ a2= 9
Length of conjugate axis = 2b
Given, length of conjugate axis = 9
⇒ 2b = 9
⇒ b =
⇒ b2=
The required equation of hyperbola is

i.e.,

(ix) Let the required equation of hyperbola be

Length of transverse axis = 2a
Given, length of transverse axis = 8
⇒ 2a = 8
⇒ a = 4
⇒ a2= 16
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a2e2= 25
Now, b2= a2(e2– 1)
⇒ b2= a2e2– a2
⇒ b2= 25 – 16 = 9
The required equation of hyperbola is

Question 6 Maharashtra Board Solution
Find the equation of the tangent to the hyperbola. (i) 3x2 – y2 = 4 at the point (2, 2√2). (ii) 3x2 – y2 = 12 at the point (4, 6) (iii) at the point whose eccentric angle is . (iv) at the point in a first quadrant whose ordinate is 3. (v) 9x2 – 16y2 = 144 at the point L of the latus rectum in the first quadrant.
Solution & Step-by-Step Answer:

Question 7 Maharashtra Board Solution
Show that the line 3x – 4y + 10 = 0 is a tangent to the hyperbola x2 – 4y2 = 20. Also, find the point of contact.
Solution & Step-by-Step Answer:
Given equation of the hyperbola is x2 – 4y2 = 20 Comparing this equation with , we get a2 = 20 and b2 = 5 Given equation of line is 3x – 4y + 10 = 0. 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 hyperbola , we must have

Question 8 Maharashtra Board Solution
If the line 3x – 4y = k touches the hyperbola , then find the value of k.
Solution & Step-by-Step Answer:
Given equation of the hyperbola is Comparing this equation with , we get a2 = 5, b2 = Given equation of line is 3x – 4y = k y = Comparing this equation with y = mx + c, we get m = , c = For the line y = mx + c to be a tangent to the hyperbola , we must have c2 = a2 m2 – b2 ⇒ ⇒ ⇒ ⇒ k2 = 25 ⇒ k = ±5

Alternate method:
Given equation of the hyperbola is
…….(i)
Given equation of the line is 3x – 4y = k
y =
Substituting this value ofy in (i), we get

⇒
⇒ 4x2– (9x2– 6kx + k2) = 20
⇒ 4x2– 9x2+ 6kx – k2= 20
⇒ -5x2+ 6kx – k2= 20
⇒ 5x2– 6kx + (k2+ 20) = 0 …..(ii)
Since, the given line touches the given hyperbola.
The quadratic equation (ii) in x has equal roots.
(-6k)2– 4(5)(k2+ 20) = 0
⇒ 36k2– 20k2– 400 = 0
⇒ 16k2= 400
⇒ k2= 25
⇒ k = ±5

Question 9 Maharashtra Board Solution
Find the equations of the tangents to the hyperbola making equal intercepts on the co-ordinate axes.
Solution & Step-by-Step Answer:
Given equation of the hyperbola is . Comparing this equation with , we get a2 = 25 and b2 = 9 Since the tangents make equal intercepts on the co-ordinate axes, ∴ m = -1 Equations of tangents to the hyperbola having slope m are y = mx ± ⇒ y = -x ± ⇒ y = -x ± √16 ⇒ x + y = ±4
Question 10 Maharashtra Board Solution
Find the equations of the tangents to the hyperbola 5x2 – 4y2 = 20 which are parallel to the line 3x + 2y + 12 = 0.
Solution & Step-by-Step Answer:
Given equation of the hyperbola is 5x2 – 4y2 = 20 Comparing this equation with , we get a2 = 4 and b2 = 5 Slope of the line 3x + 2y + 12 = 0 is Since the given line is parallel to the tangents, Slope of the required tangents (m) = Equations of tangents to the hyperbola having slope m are y = mx ±

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