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

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

Question 1.

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) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=1\)

(ii) \(\frac{x^{2}}{25}-\frac{y^{2}}{16}=-1\)

(iii) 16x^{2} – 9y^{2} = 144

(iv) 21x^{2} – 4y^{2} = 84

(v) 3x^{2} – y^{2} = 4

(vi) x^{2} – y^{2} = 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

a^{2} = 25 and b^{2} = 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

b^{2} = 16 and a^{2} = 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 16x^{2} – 9y^{2} = 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

a^{2} = 9 and b^{2} = 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 21x^{2} – 4y^{2} = 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

a^{2} = 4 and b^{2} = 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 3x^{2} – y^{2} = 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

a^{2} = \(\frac{4}{3}\) and b^{2} = 4

(vi) Given equation of the hyperbola is x^{2} – y^{2} = 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

a^{2} = 16 and b^{2} = 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

b^{2} = 25 and a^{2} = 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

b^{2} = 25 and a^{2} = 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

a^{2} = 100 and b^{2} = 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 sec^{2} θ – tan^{2} 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

a^{2} = 4 and b^{2} = 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

⇒ b^{2} = 25

Co-ordinates of focus are (-ae, 0)

ae = 7

⇒ a^{2}e^{2} = 49

Now, b^{2} = a^{2}(e^{2} – 1)

⇒ 25 = 49 – a^{2}

⇒ a^{2} = 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 x^{2} – 3y^{2} = 3

Solution:

Given, equation of hyperbola is x^{2} – 3y^{2} = 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

b^{2} = 1 and a^{2} = 3

Now, a^{2} = b^{2}(e^{2} – 1)

⇒ 3 = 1(e^{2} – 1)

⇒ 3 = e – 1

⇒ e^{2} = 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

⇒ a^{2} = 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

⇒ b^{2} = 9

Distance between foci = 2ae

Given, distance between foci = 10

⇒ 2ae = 10

⇒ ae = 5

⇒ a^{2}e^{2} = 25

Now, b^{2} = a^{2} (e^{2} – 1)

⇒ b^{2} = a^{2} e^{2} – a^{2}

⇒ 9 = 25 – a^{2}

⇒ a^{2} = 25 – 9

⇒ a^{2} = 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)

⇒ b^{2} = 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 = \(\frac{4}{3}\)

⇒ a^{2} = \(\frac{16}{9}\)

(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

⇒ a^{2} = 9

Length of conjugate axis = 2b

Given, length of conjugate axis = 9

⇒ 2b = 9

⇒ b = \(\frac{9}{2}\)

⇒ b^{2} = \(\frac{81}{4}\)

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

⇒ a^{2} = 16

Distance between foci = 2ae

Given, distance between foci = 10

⇒ 2ae = 10

⇒ ae = 5

⇒ a^{2}e^{2} = 25

Now, b^{2} = a^{2} (e^{2} – 1)

⇒ b^{2} = a^{2} e^{2} – a^{2}

⇒ b^{2} = 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) 3x^{2} – y^{2} = 4 at the point (2, 2√2).

(ii) 3x^{2} – y^{2} = 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) 9x^{2} – 16y^{2} = 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 x^{2} – 4y^{2} = 20. Also, find the point of contact.

Solution:

Given equation of the hyperbola is x^{2} – 4y^{2} = 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

a^{2} = 20 and b^{2} = 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

a^{2} = 5, b^{2} = \(\frac{5}{4}\)

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

c^{2} = a^{2} m^{2} – b^{2}

⇒ \(\left(\frac{-\mathrm{k}}{4}\right)^{2}=5\left(\frac{3}{4}\right)^{2}-\frac{5}{4}\)

⇒ \(\frac{\mathrm{k}^{2}}{16}=\frac{5}{16}(9-4)\)

⇒ \(\frac{\mathrm{k}^{2}}{16}=\frac{5}{16}(5)\)

⇒ k^{2} = 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

\(\frac{x^{2}}{5}-\frac{4}{5}\left(\frac{3 x-\mathrm{k}}{4}\right)^{2}=1\)

⇒ \(\frac{x^{2}}{5}-\frac{4}{5}\left(\frac{9 x^{2}-6 k x+k^{2}}{16}\right)=1\)

⇒ 4x^{2} – (9x^{2} – 6kx + k^{2}) = 20

⇒ 4x^{2} – 9x^{2} + 6kx – k^{2} = 20

⇒ -5x^{2} + 6kx – k^{2} = 20

⇒ 5x^{2} – 6kx + (k^{2} + 20) = 0 …..(ii)

Since, the given line touches the given hyperbola.

The quadratic equation (ii) in x has equal roots.

(-6k)^{2} – 4(5)(k^{2} + 20) = 0

⇒ 36k^{2} – 20k^{2} – 400 = 0

⇒ 16k^{2} = 400

⇒ k^{2} = 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

a^{2} = 25 and b^{2} = 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 5x^{2} – 4y^{2} = 20 which are parallel to the line 3x + 2y + 12 = 0.

Solution:

Given equation of the hyperbola is 5x^{2} – 4y^{2} = 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

a^{2} = 4 and b^{2} = 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 ± \(\sqrt{a^{2} m^{2}-b^{2}}\)