Students must practice this TS Intermediate Maths 2B Solutions Chapter 5 Hyperbola Ex 5(a) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2B Solutions Chapter 5 Hyperbola Exercise 5(a)

I.

Question 1.

One focus of a hyperbola is located at the point (1, -3) and the corresponding directrix is the line y = 2. Find the equation of the hyperbola if its eccentricity is \(\frac{3}{2}\). (May 2009)

Solution:

Given one focus of hyperbola is at (1, -3)

∴ S = (1, -3) and directrix is y – 2 = 0

Given e = \(\frac{3}{2}\).

Let P(x_{1}, y_{1}) be a point on the locus. Then

SP = e . PM

⇒ SP^{2} = e^{2} . PM^{2}

∴ Locus of (x_{1}, y_{1}) is the equation of hyperbola 4x^{2} – 5y^{2} – 8x + 60y + 4 = 0.

Question 2.

If the lines 3x – 4y = 12 and 3x + 4y = 12 meets on a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) then find the eccentricity of the hyperbola.

Solution:

Given the equation of lines are

3x – 4y = 12 …….(1)

3x + 4y = 12 ………(2)

The combined equations of lines (1) and (2) is (3x – 4y) (3x + 4y) = 144

⇒ 9x^{2} – 16y^{2} = 144

⇒ \(\frac{x^2}{16}-\frac{y^2}{9}=1\) which represents a hyperbola of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\).

Question 3.

Find the equations of hyperbola whose foci are (±5, 0); the transverse axis is of length 8.

Solution:

Given foci as (±5, 0)and comparing with the standard equation we get

ae = 5 …….(1)

and the length of the transverse axis is 2a = 8

⇒ a = 4

∴ ae = 5

⇒ e = \(\frac{5}{4}\)

since b^{2} = a^{2}(e^{2} – 1)

⇒ b^{2} = 16\(\left(\frac{25}{16}-1\right)\) = 9

∴ Equation of hyperbola whose foci are (±5, 0) and the transverse axis of length ‘8’ is \(\frac{x^2}{16}-\frac{y^2}{9}=1\)

⇒ 9x^{2} – 16y^{2} = 144

Question 4.

Find the equation of hyperbola whose asymptotes are the straight lines (x + 2y + 3) = 0 and (3x + 4y + 5) = 0 and which passes through the point (1, -1).

Solution:

Given asymptotes are x + 2y + 3 = 0 and 3x + 4y + 5 = 0.

∴ The equation of point of asymptotes is (x + 2y + 3) (3x + 4y + 5) = 0

⇒ 3x^{2} + 6xy + 9x + 4xy + 8y^{2} + 10y + 9x + 12y + 15 = 0

⇒ 3x^{2} + 10xy + 8y^{2} + 14x + 22y + 15 = 0

∴ The equation of hyperbola having the given lines as asymptotes are

3x^{2} + 10xy + 8y^{2} + 14x + 22y + k = 0 ……..(1)

Given (1) is passing through (1, -1) then

3 – 10 + 8 + 14 – 22 + k = 0

⇒ k = 7

∴ From (1) the equation of the required hyperbola is 3x^{2} + 10xy + 8y^{2} + 14x + 22y + 7 = 0

Question 5.

If 3x – 4y + k = 0 is a tangent to x^{2} – 4y^{2} = 5, find the value of k.

Solution:

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

⇒ \(\frac{x^2}{5}-\frac{y^2}{\left(\frac{5}{4}\right)}=1\)

Comparing with \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

We get a^{2} = 5 and b^{2} = \(\frac{5}{4}\)

Given 3x – 4y + k = 0

⇒ 4y = 3x + 3

⇒ y = \(\frac{3}{4} x+\frac{k}{4}\)

Question 6.

Find the product of lengths of perpendiculars from any point on the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) to its asymptotes.

Solution:

Let S ≡ \(\frac{x^2}{16}-\frac{y^2}{9}-1=0\) be the given hyperbola.

Let P = (a sec θ, b tan θ) be any point on S = 0.

The equation of asymptotes of the hyperbola S = 0 are \(\frac{x}{4}+\frac{y}{3}=0\) and \(\frac{x}{4}-\frac{y}{3}=0\)

⇒ 3x + 4y = 0 ………(1) and 3x – 4y = 0 ………(2)

Let PM be the length of the perpendicular drawn from P(4 sec θ, 3 tan θ) to the line (1) then

PM = \(\left|\frac{12 \sec \theta+12 \tan \theta}{\sqrt{9+16}}\right|\) = \(\left|\frac{12 \sec \theta+12 \tan \theta}{5}\right|\)

Let PN be the length of the perpendicular from P(4 sec θ, 3 tan θ) on line (2). Then

PN = \(\frac{|12 \sec \theta-12 \tan \theta|}{\sqrt{9+16}}=\frac{|12 \sec \theta-12 \tan \theta|}{5}\)

∴ Product of perpendiculars = PM . PN

= \(\frac{|12 \sec \theta+12 \tan \theta|}{5} \cdot \frac{|12 \sec \theta-12 \tan \theta|}{5}\)

= \(\frac{144\left(\sec ^2 \theta-\tan ^2 \theta\right)}{25}\)

= \(\frac{144}{25}\)

∴ Product of length of perpendiculars = \(\frac{144}{25}\)

Question 7.

If the eccentricity of a hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate hyperbola. (March 2012, 2013)

Solution:

If S ≡ \(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1=0\) is the equation of hyperbola then S’ ≡ \(\frac{x^2}{a^2}-\frac{y^2}{b^2}+1=0\) is the equation of the conjugate hyperbola.

Let e and e_{1} be the eccentricities of the hyperbola and its conjugate respectively,

∴ The eccentricity of the conjugate hyperbola is \(\frac{5}{3}\).

Question 8.

Find the equation of the hyperbola whose asymptotes are 3x = ±5y and the vertices and (±5, 0).

Solution:

The equation of asymptotes is given by 3x – 5y = 0 and 3x + 5y = 0.

∴ The equation of hyperbola is of the form (3x – 5y) (3x + 5y) = k

⇒ 9x^{2} – 25y^{2} = k

If the hyperbola passes through the vertex (±5, 0) then 9(25) = k

⇒ k = 225

Hence the equation of asymptotes of a hyperbola is 9x^{2} – 25y^{2} = 225

Question 9.

Find the equation of normal at θ = \(\frac{\pi}{3}\) to the hyperbola 3x^{2} – 4y^{2} = 12.

Solution:

The given equation of the hyperbola is 3x^{2} – 4y^{2} = 12

⇒ \(\frac{x^2}{4}-\frac{y^2}{3}=1\)

The equation of normal at P(a sec θ, b tan θ) to the hyperbola S = 0 is

Question 10.

If the angle between asymptotes is 30° then find its eccentricity.

Solution:

The angle between asymptotes of the hyperbola

II.

Question 1.

Find the centre, foci, eccentricity, equation of directrices, and length of the latus rectum of the following hyperbolas.

(i) 16y^{2} – 9x^{2} = 144 (June 2010)

Solution:

The given equation of the hyperbola is 16y^{2} – 9x^{2} = 144

⇒ \(\frac{y^2}{9}-\frac{x^2}{16}=1\)

⇒ \(\frac{x^2}{16}-\frac{y^2}{9}=-1\)

Comparing with \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1\) We get

a^{2} = 16 and b^{2} = 9

⇒ a = 4 and b = 3

(i) Centre of the hyperbola = (0, 0)

(ii) x^{2} – 4y^{2} = 4 (May 2011)

Solution:

This can be written as \(\frac{x^2}{4}-\frac{y^2}{1}=1\)

⇒ a^{2} = 4 and b^{2} = 1

⇒ a = 2 and b = 1

(i) Centre = (0, 0)

(iii) 5x^{2} – 4y^{2} + 20x + 8y – 4 = 0 (Mar 2012)

Solution:

The given equation is 5x^{2} – 4y^{2} + 20x + 8y – 4 = 0

⇒ 5x^{2} + 20x – 4y^{2} + 8y = 4

⇒ 5(x^{2} + 4x) – 4(y^{2} – 2y) = 4

⇒ 5(x^{2} + 4x + 4) – 4(y^{2} – 2y + 1) = 20 + 4 – 4 = 20

⇒ 5(x + 2)^{2} – 4(y – 1)^{2} = 20

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

This is of the form \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

Where a^{2} = 4 and b^{2} = 5, h = -2, k = 1

(i) Centre of the hyperbola = C(h, k) = (-2, 1)

(ii) Eccentricity = \(\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{4+5}{4}}=\frac{3}{2}\)

(iii) Foci = (h ± ae, k)

= (-2 ± 2(\(\frac{3}{2}\)), 1)

= (1, 1), (-5, 1)

∴ Foci of the hyperbola = (1, 1), (-5, 1)

(iv) Equations of directrices are x = h ± \(\frac{a}{e}\)

= \(-2 \pm \frac{2}{(3 / 2)}=-2 \pm \frac{4}{3}\)

∴ 3x = -6 ± 4

⇒ 3x = -2 and 3x = -10

⇒ 3x + 2 = 0 and 3x + 10 = 0

(v) Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 5}{2}\) = 5

(iv) 9x^{2} – 16y^{2} + 72x – 32y – 16 = 0

Solution:

The given equation is 9x^{2} – 16y^{2} + 72x – 32y – 16 = 0

⇒ 9x^{2} + 72x – 16y^{2} – 32y = 16

⇒ 9(x^{2} + 8x) – 16(y^{2} + 2y) = 16

⇒ 9(x^{2} + 8x + 16) – 16(y^{2} + 2y + 1) = 16 + 144 – 16 = 144

⇒ \(\frac{(x-(-4))^2}{16}+\frac{[y-(-1)]^2}{9}=1\)

This is of the form \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\)

Where a^{2} = 16 and b^{2} = 9

⇒ a = 4 and b = 3

(i) Coordinates of centre = C(h, k) = C(-4, -1)

(ii) Eccentricity e = \(\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{16+9}{16}}=\frac{5}{4}\)

(iii) Coordinates of foci = (h ± ae, k)

= (-4 ± 4(\(\frac{5}{4}\)), -1)

= (-9, -1), (1, -1)

(iv) Equations of directrices are x = ±\(\frac{a}{e}\)

= \(-4 \pm \frac{4}{(5 / 4)}\)

= \(-4 \pm \frac{16}{5}\)

⇒ x + 4 = ±\(\frac{16}{5}\)

⇒ 5x + 20 = 16 or 5x + 20 = -16

⇒ 5x + 4 = 0 (or) 5x + 36 = 0

(v) Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2(9)}{4}=\frac{9}{2}\)

Question 2.

Find the equation to the hyperbola whose foci are (4, 2) and (8, 2) and whose eccentricity is 2. (March 2009)

Solution:

Let S = (4, 2) and S’ = (8, 2) be the foci.

Since the Y-coordinate of S and S’ are the same, the hyperbola’s major axis is parallel to the X-axis.

The equation of required hyperbola is \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

Centre of Hyperbola C = Midpoint of SS’

= \(\left(\frac{4+8}{2}, \frac{2+2}{2}\right)\)

= (2, 2)

Given that e = 2

Distance between the foci is SS’ = 2ae

∴ 2ae = \(\sqrt{(4-8)^2+(2-2)^2}\) = 4

⇒ ae = 2

⇒ a(2) = 2

⇒ a = 1

But b^{2} = a^{2}(e^{2} – 1) = 1(4 – 1) = 3

The equation of required hyperbola is \(\frac{(x-6)^2}{1}-\frac{(y-2)^2}{3}=1\)

⇒ (x^{2} – 12x + 36) – \(\frac{\left(y^2-4 y+4\right)}{3}\) = 1

⇒ 3x^{2} – y2 – 36x + 4y + 101 = 0

∴ Equation of Hyperbola is 3x^{2} – y^{2} – 36x + 4y + 101 = 0

Question 3.

Find the equation of hyperbola of a given length of transverse axis 6 whose vertex bisects the distance between the centre and the focus.

Solution:

Let C be the centre.

S is the focus and A, A’ are vertices of the required hyperbola.

C = (0, 0), S = (ae, 0);

Given AA’ = 2a = 6

⇒ CA = a = 3; A = (3, 0)

Given that A bisects \(\overline{\mathrm{CS}}\)

∴ A is midpoint of \(\overline{\mathrm{CS}}\)

∴ (\(\frac{ae}{2}\)) = a

⇒ e = 2

But b^{2} = a^{2}(e^{2} – 1)

⇒ b^{2} = a^{2}(4 – 1) = 3a^{2} = 27

∴ The equation of the required hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

⇒ \(\frac{x^2}{9}-\frac{y^2}{27}=1\)

⇒ 3x^{2} – y^{2} = 27

Question 4.

Find the equation of the tangents to the hyperbola x^{2} – 4y^{2} = 4 which are (i) parallel (ii) perpendicular to the line x + 2y = 0. (May 2011)

Solution:

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

⇒ \(\frac{x^2}{4}-\frac{y^2}{1}=1\) ……..(1)

∴ a^{2} = 4, b^{2} = 1

Given line is x + 2y = 0 ……..(2)

(i) Equation of any line parallel to (2) is x + 2y + k = 0

⇒ 2y = -x – k

⇒ y = \(-\frac{x}{2}-\frac{k}{2}\) ……(3)

Condition for (3) to be a tangent to the hyperbola (1) is c^{2} = a^{2}m^{2} – b^{2}

where c = \(-\frac{k}{2}\), m = \(-\frac{1}{2}\)

∴ \(\frac{k^2}{4}=4\left(\frac{1}{4}\right)-1\)

⇒ \(\frac{k^2}{4}\) = 0

⇒ k = 0

The equation of tangent parallel to x + 2y = 0 is x + 2y = 0.

(ii) Equation of line perpendicular to x + 2y = 0 is 2x – y + k = 0.

⇒ y = 2x + k, where m = 2 and c = k

∴ c^{2} = a^{2}m^{2} – b^{2}

⇒ k^{2} = 4(4) – 1 = 15

⇒ k = ±√15

∴ Equation of tangent perpendicular to x + 2y = 0 is 2x – y ± √15 = 0.

Question 5.

Find the equations of tangents drawn to the hyperbola 2x^{2} – 3y^{2} = 6 through (-2, 1).

Solution:

Given equation of hyperbola 2x^{2} – 3y^{2} = 6

⇒ \(\frac{x^2}{3}-\frac{y^2}{2}=1\) …….(1)

Comparing (1) with \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) we get a^{2} = 3, b^{2} = 2

Equation of any tangent to the hyperbola (1) having slope ‘m’ is y = mx ± \(\sqrt{a^2 m^2-b^2}\)

⇒ y = mx ± \(\sqrt{3 m^2-2}\) ……….(2)

If the tangent (2) passes through (-2, 1) then 1 = -2m ± \(\sqrt{3 m^2-2}\)

⇒ (2m + 1)^{2} = 3m^{2} – 2

⇒ 4m^{2} + 4m + 1 = 3m^{2} – 2

⇒ m^{2} + 4m + 3 = 0

⇒ (m + 3) (m + 1) = 0

⇒ m = -1 (or) m = -3

∴ The equations of tangents from (2) are y = -x ± √1 and y = -3x ± √25 = -3x ± 5

∴ x + y ± 1 = 0 and 3x + y ± 5 = 0 are the equations.

But the point (-2, 1) does not satisfy x + y – 1 = 0 and 3x + y – 5 = 0.

Hence the equations of required tangents passing through (-2, 1) are x + y + 1 = 0 and 3x + y + 5 = 0.

Question 6.

Prove that the product of the perpendicular distances from any point on a hyperbola to its asymptotes is constant.

Solution:

Let S ≡ \(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0\) be the given hyperbola.

Let P = (a sec θ, b tan θ) be any point on S = 0.

The equations of asymptotes of hyperbola S = 0 are \(\frac{x}{a}+\frac{y}{b}=0\) and \(\frac{x}{a}-\frac{y}{b}=0\)

⇒ bx + ay = 0 (1) and bx – ay = 0 ………(2)

Let PM be the perpendicular length drawn from P(a sec θ, b tan θ) on line (2).

∴ PM = \(\frac{|b a \sec \theta+a b \tan \theta|}{\sqrt{a^2+b^2}}\)

Let PN be the perpendicular length drawn from P(a sec θ, b tan θ) on line (2).

∴ The product of the perpendicular distances from any point on a hyperbola to its asymptotes is a constant.

III.

Question 1.

Tangents to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) make angles θ1, θ2 with transverse axis of a hyperbola. Show that the point of intersection of these tangents lies on the curve 2xy = k(x^{2} – a^{2}) when tan θ1 + tan θ2 = k.

Solution:

Given hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

The transverse axis of the hyperbola is the x-axis, (i.e., y = 0)

Let P(x_{1}, y_{1}) be the point of intersection of the tangents drawn to the given hyperbola.

The equation of any tangent to the hyperbola is of the form y = mx ± \(\sqrt{a^2 m^2-b^2}\)

If this passes through (x_{1}, y_{1}) then

This is a quadratic equation in ‘m’ and be m_{1}, m_{2} be the roots which corresponds to the slopes tan θ_{1}, tan θ_{2} of tangents.

∴ Locus of (x_{1}, y_{1}) is the curve 2xy = k(x^{2} – a^{2}).

Question 2.

Show that the locus of feet of the perpendiculars drawn from foci to any tangent of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is the auxiliary circle of the hyperbola.

Solution:

Let S ≡ \(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1=0\) be the given hyperbola.

Foci of the hyperbola S = 0 are (±ae, 0).

Equation of any tangent to S = 0 having slope ‘m’ is y = mx ± \(\sqrt{a^2 m^2-b^2}\)

⇒ y – mx = ±\(\sqrt{a^2 m^2-b^2}\) ………(1)

Let P(x_{1}, y_{1}) be the locus of feet of the perpendicular form foci of the hyperbola to the tangent (1).

The equation of the perpendicular from either focus (±ae, 0) on the above tangent is

y – 0 = \(-\frac{1}{m}\)(x – (±ae))

⇒ my + x = ±ae …….(2)

P(x_{1}, y_{1}) lies on (1) and (2).

y_{1} – mx_{1} = \(\pm \sqrt{a^2 m^2-b^2}\) …….(3)

and my_{1} + x_{1} = ±ae ………(4)

Eliminating m from equations (3) and (4)

∴ The Locus of (x_{1}, y_{1}) is x^{2} + y^{2} = a^{2} which is the equation of the Auxiliary circle of the Hyperbola.

Question 3.

Show that the equation \(\frac{x^2}{9-c}+\frac{y^2}{5-c}=1\) represents

(i) an ellipse if ‘c’ is a real constant less than 5.

(ii) a hyperbola if ‘c’ is any real constant between 5 and 9.

(iii) show that each ellipse in (i) and each hyperbola (ii) has foci at the two points (±2, 0), independent of the value of ‘c’.

Solution:

Given equation \(\frac{x^2}{9-c}+\frac{y^2}{5-c}=1\) ……(1) represents an ellipse

if 9 – c > 0 and 5 – c > 0

⇒ 9 > c and 5 > c

⇒ c < 9 and c < 5

⇒ c < 5 ∴ c is a real constant and less than 5 if (1) represents an ellipse.

(i) The equation of the hyperbola is of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and the given equation (1) represents a hyperbola

if 9 – c > 0 and 5 – c < 0

⇒ 9 > c and 5 < c

⇒ 5 < c < 9

∴ (1) represents hyperbola if C is a real constant such that 5 < c < 9

(ii) If \(\frac{x^2}{9-c}+\frac{y^2}{5-c}=1\) represents ellipse then

a^{2} = 9 – c and b^{2} = 5 – c

Eccentricity b^{2} = a^{2}(1 – e^{2})

⇒ 5 – c = (9 – c) (1 – e^{2})

Hence each ellipse in (i) and each hyperbola in (ii) has foci at the two points (±2, 0) independent of the value of C.

Question 4.

Show that the angle between the two asymptotes of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is \(2 {tan}^{-1}\left(\frac{b}{a}\right)\) (or) 2 sec-1(e). [New Model Paper, May 2012]

Solution:

Let the equation of the hyperbola be \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

The asymptotes of hyperbola are y = ±\(\frac{b}{a}\)x where m1 = \(\frac{b}{a}\) and m2 = \(-\frac{b}{a}\).

If θ is the angle between asymptotes of the hyperbola then