Conic Sections Class 11 Maths 1 Exercise 7.3 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.3 Questions and Answers.

11th Maths Part 1 Conic Sections Exercise 7.3 Questions And Answers Maharashtra Board

Question 1.
Find the length of the transverse axis, length of the conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices, and the length of the latus rectum of the hyperbolae.
(i) $\frac{x^{2}}{25}-\frac{y^{2}}{16}=1$
(ii) $\frac{x^{2}}{25}-\frac{y^{2}}{16}=-1$
(iii) 16x² – 9y² = 144
(iv) 21x² – 4y² = 84
(v) 3x² – y² = 4
(vi) x² – y² = 16
(vii) $\frac{y^{2}}{25}-\frac{x^{2}}{9}=1$
(viii) $\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$
(ix) $\frac{x^{2}}{100}-\frac{y^{2}}{25}=1$
(x) x = 2 sec θ, y = 2√3 tan θ
Solution:
(i) Given equation of the hyperbola is $\frac{x^{2}}{25}-\frac{y^{2}}{16}=1$
Comparing this equation with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, 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 $\frac{x^{2}}{25}-\frac{y^{2}}{16}=-1$
$\frac{y^{2}}{16}-\frac{x^{2}}{25}=1$
Comparing this equation with $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$, 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.
$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$
Comparing this equation with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, 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 = $\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{9+16}}{3}=\frac{\sqrt{25}}{3}=\frac{5}{3}$
Co-ordinates of foci are S(ae, 0) and S'(-ae, 0),
i.e., S(3($\frac{5}{3}$), 0) and S'(-3($\frac{5}{3}$), 0)
i.e., S(5, 0) and S'(-5, 0)
Equations of the directrices are x = ±$\frac{a}{e}$
= $\pm \frac{3}{\left(\frac{5}{3}\right)}$
= $\pm \frac{9}{5}$
Length of latus rectum = $\frac{2 b^{2}}{a}=\frac{2(16)}{3}=\frac{32}{3}$

(iv) Given equation of the hyperbola is 21x² – 4y² = 84.
$\frac{x^{2}}{4}-\frac{y^{2}}{21}=1$
Comparing this equation with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, 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.
$\frac{x^{2}}{\left(\frac{4}{3}\right)}-\frac{y^{2}}{4}=1$
Comparing this equation with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we get
a² = $\frac{4}{3}$ and b² = 4

(vi) Given equation of the hyperbola is x² – y² = 16.
$\frac{x^{2}}{16}-\frac{y^{2}}{16}=1$
Comparing this equation with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, 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 = $\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{16+16}}{4}=\frac{\sqrt{32}}{4}=\frac{4 \sqrt{2}}{4}=\sqrt{2}$
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 = ±$\frac{a}{e}$
⇒ x = $\pm \frac{4}{\sqrt{2}}$
⇒ x = ± 2√2
Length of latus rectum = $\frac{2 b^{2}}{a}=\frac{2(16)}{4}$ = 8

(vii) Given equation of the hyperbola is $\frac{y^{2}}{25}-\frac{x^{2}}{9}=1$.
Comparing this equation with $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$, 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 $\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$.
Comparing this equation with $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$, 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 $\frac{x^{2}}{100}-\frac{y^{2}}{25}=1$
Comparing this equation with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, 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 sec² θ – tan² θ = 1,
$\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{2 \sqrt{3}}\right)^{2}=1$
$\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$
Comparing this equation with $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$, 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 = $\frac{\sqrt{a^{2}+b^{2}}}{a}$ = $\frac{\sqrt{4+12}}{2}$ = 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 = ±$\frac{a}{e}$.
⇒ x = ±$\frac{2}{2}$
⇒ x = ±1
Length of latus rectum = $\frac{2 b^{2}}{a}=\frac{2(12)}{2}$ = 12

Question 2.
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:
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 $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
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 $\frac{x^{2}}{24}-\frac{y^{2}}{25}=1$

Question 3.
Find the eccentricity of the hyperbola, which is conjugate to the hyperbola x² – 3y² = 3
Solution:
Given, equation of hyperbola is x² – 3y² = 3.
$\frac{x^{2}}{3}-\frac{y^{2}}{1}=1$
Equation of the hyperbola conjugate to the above hyperbola is $\frac{y^{2}}{1}-\frac{x^{2}}{3}=1$
Comparing this equation with $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$, 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.
If e and e’ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that $\frac{1}{e^{2}}+\frac{1}{\left(e^{\prime}\right)^{2}}=1$.
Solution:
Let e be the eccentricity of a hyperbola

Question 5.
Find the equation of the hyperbola referred to its principal axes:
(i) whose distance between foci is 10 and eccentricity is $\frac{5}{2}$
(ii) whose distance between foci is 10 and length of the conjugate axis is 6.
(iii) whose distance between directrices is $\frac{8}{3}$ and eccentricity is $\frac{3}{2}$.
(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 $\frac{3}{2}$.
(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:
(i) Let the required equation of hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Given, eccentricity (e) = $\frac{5}{2}$
Distance between foci = 2ae
Given, distance between foci = 10
⇒ 2ae = 10
⇒ ae = 5
⇒ a($\frac{5}{2}$) = 5
⇒ a = 2
⇒ a² = 4

(ii) Let the required equation of hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
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 = a² e² – a²
⇒ 9 = 25 – a²
⇒ a² = 25 – 9
⇒ a² = 16
The required equation of hyperbola is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$

(iii) Let the required equation of hyperbola be $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$
Given, eccentricity (e) = $\frac{3}{2}$
Distance between directrices = $\frac{2a}{e}$
Given, distance between directrices = $\frac{8}{3}$

(iv) Let the required equation of hyperbola be
$\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$ ……(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
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ……(i)
The hyperbola passes through the points (6, 9) and (3, 0).

(vi) Let the required equation of hyperbola be
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
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 $\frac{x^{2}}{7^{2}}-\frac{y^{2}}{3^{2}}=1$
i.e., $\frac{x^{2}}{49}-\frac{y^{2}}{9}=1$

(vii) Let the required equation of hyperbola be
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ……(i)
Given, eccentricity (e) = $\frac{3}{2}$
Co-ordinates of foci are (±ae, 0).
Given co-ordinates of foci are (±2, 0)
ae = 2
⇒ a($\frac{3}{2}$) = 2
⇒ a = $\frac{4}{3}$
⇒ a² = $\frac{16}{9}$

(viii) Let the required equation of hyperbola be
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
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 = $\frac{9}{2}$
⇒ b² = $\frac{81}{4}$
The required equation of hyperbola is
$\frac{x^{2}}{9}-\frac{y^{2}}{\left(\frac{81}{4}\right)}=1$
i.e., $\frac{x^{2}}{9}-\frac{4 y^{2}}{81}=1$

(ix) Let the required equation of hyperbola be
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
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 = a² e² – a²
⇒ b² = 25 – 16 = 9
The required equation of hyperbola is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$

Question 6.
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) $\frac{x^{2}}{144}-\frac{y^{2}}{25}=1$ at the point whose eccentric angle is $\frac{\pi}{3}$.
(iv) $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ 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:

Question 7.
Show that the line 3x – 4y + 10 = 0 is a tangent to the hyperbola x² – 4y² = 20. Also, find the point of contact.
Solution:
Given equation of the hyperbola is x² – 4y² = 20
$\frac{x^{2}}{20}-\frac{y^{2}}{5}=1$
Comparing this equation with $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$, we get
a² = 20 and b² = 5
Given equation of line is 3x – 4y + 10 = 0.
y = $\frac{3 x}{4}+\frac{5}{2}$
Comparing this equation with y = mx + c, we get
m = $\frac{3}{4}$ and c = $\frac{5}{2}$
For the line y = mx + c to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we must have

Question 8.
If the line 3x – 4y = k touches the hyperbola $\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1$, then find the value of k.
Solution:
Given equation of the hyperbola is
$\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1$
$\frac{x^{2}}{5}-\frac{y^{2}}{\frac{5}{4}}=1$
Comparing this equation with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we get
a² = 5, b² = $\frac{5}{4}$
Given equation of line is 3x – 4y = k
y = $\frac{3}{4} x-\frac{\mathrm{k}}{4}$
Comparing this equation with y = mx + c, we get
m = $\frac{3}{4}$, c = $-\frac{\mathrm{k}}{4}$
For the line y = mx + c to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we must have
c² = a² m² – b²
⇒ $\left(\frac{-\mathrm{k}}{4}\right)^{2}=5\left(\frac{3}{4}\right)^{2}-\frac{5}{4}$
⇒ $\frac{\mathrm{k}^{2}}{16}=\frac{5}{16}(9-4)$
⇒ $\frac{\mathrm{k}^{2}}{16}=\frac{5}{16}(5)$
⇒ k² = 25
⇒ k = ±5

Alternate method:
Given equation of the hyperbola is
$\frac{x^{2}}{5}-\frac{4 y^{2}}{5}=1$ …….(i)
Given equation of the line is 3x – 4y = k
y = $\frac{3 x-\mathrm{k}}{4}$
Substituting this value ofy in (i), we get
$\frac{x^{2}}{5}-\frac{4}{5}\left(\frac{3 x-\mathrm{k}}{4}\right)^{2}=1$
⇒ $\frac{x^{2}}{5}-\frac{4}{5}\left(\frac{9 x^{2}-6 k x+k^{2}}{16}\right)=1$
⇒ 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.
Find the equations of the tangents to the hyperbola $\frac{x^{2}}{25}-\frac{y^{2}}{9}=1$ making equal intercepts on the co-ordinate axes.
Solution:
Given equation of the hyperbola is $\frac{x^{2}}{25}-\frac{y^{2}}{9}=1$.
Comparing this equation with $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$, 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 $\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1$ having slope m are
y = mx ± $\sqrt{a^{2} m^{2}-b^{2}}$
⇒ y = -x ± $\sqrt{25(-1)^{2}-9}$
⇒ y = -x ± √16
⇒ x + y = ±4

Question 10.
Find the equations of the tangents to the hyperbola 5x² – 4y² = 20 which are parallel to the line 3x + 2y + 12 = 0.
Solution:
Given equation of the hyperbola is 5x² – 4y² = 20
$\frac{x^{2}}{4}-\frac{y^{2}}{5}=1$
Comparing this equation with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, we get
a² = 4 and b² = 5
Slope of the line 3x + 2y + 12 = 0 is $-\frac{3}{2}$
Since the given line is parallel to the tangents,
Slope of the required tangents (m) = $-\frac{3}{2}$
Equations of tangents to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having slope m are
y = mx ± $\sqrt{a^{2} m^{2}-b^{2}}$

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1
• Exercise 8.1 Class 11 Maths 1
• Exercise 8.2 Class 11 Maths 1
• Exercise 8.3 Class 11 Maths 1
• Miscellaneous Exercise 8 Class 11 Maths 1