Differentiation (Part-38) | S N De

In this article, we have solved a few Short Answer Type questions (1-3) of Hyperbola related problems of S N Dey mathematics class 11.

1. Find (i) the length of axes (ii) length of latus rectum (iii) co-ordinates of vertices (iv) eccentricity (v) co-ordinates of foci and (vi) equations of the directrices of each of the following hyperbolas:
(a) \( 4x^2-9y^2=36 \)    (b) \( 9y^2-25x^2=225 \)

Solution (a)

\[ \frac{4x^2}{36} - \frac{9y^2}{36} = 1 \Rightarrow \frac{x^2}{9} - \frac{y^2}{4} = 1 \quad \rightarrow (1) \]

Comparing (1) with \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we get \( a^2 = 9 \Rightarrow a = 3 \) and \( b^2 = 4 \Rightarrow b = 2 \).

(i) Length of transverse axis \( = 2a = 6 \) unit, conjugate axis \( = 2b = 4 \) unit.

(ii) Length of latus rectum \( = \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3} \) unit.

(iii) Co-ordinates of vertices: \( (\pm a, 0) = (\pm 3, 0) \).

(iv) Eccentricity \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \frac{\sqrt{13}}{3} \).

(v) Co-ordinates of foci: \( (\pm ae, 0) = \left(\pm 3 \times \frac{\sqrt{13}}{3}, 0\right) = (\pm \sqrt{13}, 0) \).

(vi) Equations of directrices: \( x = \pm \frac{a}{e} = \pm \frac{3}{\sqrt{13}/3} = \pm \frac{9}{\sqrt{13}} \Rightarrow \sqrt{13}x = \pm 9 \).

Solution (b)

\[ \frac{9y^2}{225} - \frac{25x^2}{225} = 1 \Rightarrow \frac{y^2}{25} - \frac{x^2}{9} = 1 \quad \rightarrow (1) \]

Comparing (1) with \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), we get \( a^2 = 25 \Rightarrow a = 5 \) and \( b^2 = 9 \Rightarrow b = 3 \).

(i) Length of transverse axis \( = 2a = 10 \) unit, conjugate axis \( = 2b = 6 \) unit.

(ii) Length of latus rectum \( = \frac{2b^2}{a} = \frac{18}{5} \) unit.

(iii) Co-ordinates of vertices: \( (0, \pm a) = (0, \pm 5) \).

(iv) Eccentricity \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{25}} = \frac{\sqrt{34}}{5} \).

(v) Co-ordinates of foci: \( (0, \pm ae) = \left(0, \pm 5 \times \frac{\sqrt{34}}{5}\right) = (0, \pm \sqrt{34}) \).

(vi) Equations of directrices: \( y = \pm \frac{a}{e} = \pm \frac{5}{\sqrt{34}/5} = \pm \frac{25}{\sqrt{34}} \Rightarrow \sqrt{34}y = \pm 25 \).

2. Find the length of the transverse and conjugate axes of the hyperbola \( 9x^2-16y^2=144 \). Write down the equation of the hyperbola conjugate to it and find the eccentricities of both the hyperbolas.

Solution

\[ \frac{9x^2}{144} - \frac{16y^2}{144} = 1 \Rightarrow \frac{x^2}{16} - \frac{y^2}{9} = 1 \quad \rightarrow (1) \]

Comparing (1) with \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we get \( a^2 = 16 \Rightarrow a = 4 \) and \( b^2 = 9 \Rightarrow b = 3 \).

Length of transverse axis \( = 2a = 8 \) unit, conjugate axis \( = 2b = 6 \) unit.

Conjugate Hyperbola:

The equation of the conjugate hyperbola is \( \frac{y^2}{9} - \frac{x^2}{16} = 1 \Rightarrow 16y^2 - 9x^2 = 144 \).

Eccentricities:

Eccentricity of the given hyperbola \( e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4} \).

Eccentricity of the conjugate hyperbola \( e_2 = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{1 + \frac{16}{9}} = \frac{5}{3} \).

3. Find the equation of the hyperbola, whose axes are axes of co-ordinates and:

(i) length of transverse and conjugate axes are 5 and 6 respectively.

Solution

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

\( 2a = 5 \Rightarrow a = \frac{5}{2} \) and \( 2b = 6 \Rightarrow b = 3 \).

Equation: \( \frac{x^2}{(5/2)^2} - \frac{y^2}{3^2} = 1 \Rightarrow \frac{4x^2}{25} - \frac{y^2}{9} = 1 \Rightarrow 36x^2 - 25y^2 = 225 \).

(ii) lengths of conjugate axis and latus rectum are 2 and \( \frac{8}{3} \) respectively.

Solution

\( 2b = 2 \Rightarrow b = 1 \).

Latus rectum \( \frac{2b^2}{a} = \frac{8}{3} \Rightarrow \frac{2(1)^2}{a} = \frac{8}{3} \Rightarrow a = \frac{3}{4} \).

Equation: \( \frac{x^2}{(3/4)^2} - \frac{y^2}{1^2} = 1 \Rightarrow \frac{16x^2}{9} - y^2 = 1 \Rightarrow 16x^2 - 9y^2 = 9 \).

(iii) which passes through the points \( (1,1) \) and \( (2,-3) \).

Solution

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

Passes through \( (1,1) \): \( \frac{1}{a^2} - \frac{1}{b^2} = 1 \quad \rightarrow (1) \)

Passes through \( (2,-3) \): \( \frac{4}{a^2} - \frac{9}{b^2} = 1 \quad \rightarrow (2) \)

Solving (1) and (2): Multiply (1) by 4 and subtract from (2):

\( -\frac{5}{b^2} = -3 \Rightarrow b^2 = \frac{5}{3} \).

From (1): \( \frac{1}{a^2} = 1 + \frac{3}{5} = \frac{8}{5} \Rightarrow a^2 = \frac{5}{8} \).

Equation: \( \frac{x^2}{5/8} - \frac{y^2}{5/3} = 1 \Rightarrow \frac{8x^2}{5} - \frac{3y^2}{5} = 1 \Rightarrow 8x^2 - 3y^2 = 5 \).

(iv) distance between the foci is 10 and length of conjugate axis is 6.

Solution

\( 2b = 6 \Rightarrow b = 3 \).

\( 2ae = 10 \Rightarrow ae = 5 \Rightarrow a^2 e^2 = 25 \).

\( a^2\left(1 + \frac{b^2}{a^2}\right) = 25 \Rightarrow a^2 + b^2 = 25 \Rightarrow a^2 + 9 = 25 \Rightarrow a^2 = 16 \).

Equation: \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \Rightarrow 9x^2 - 16y^2 = 144 \).

(v) co-ordinates of foci are \( \left(\pm \frac{5}{2}, 0\right) \) and the length of latus rectum is \( \frac{9}{4} \).

Solution

\( ae = \frac{5}{2} \Rightarrow a^2 e^2 = \frac{25}{4} \Rightarrow a^2 + b^2 = \frac{25}{4} \quad \rightarrow (1) \)

Latus rectum \( \frac{2b^2}{a} = \frac{9}{4} \Rightarrow b^2 = \frac{9}{8}a \quad \rightarrow (2) \)

Substitute (2) in (1): \( a^2 + \frac{9}{8}a = \frac{25}{4} \Rightarrow 8a^2 + 9a - 50 = 0 \).

\( 8a^2 + 25a - 16a - 50 = 0 \Rightarrow (8a + 25)(a - 2) = 0 \).

Since \( a > 0 \), \( a = 2 \). Then \( b^2 = \frac{9}{8} \times 2 = \frac{9}{4} \).

Equation: \( \frac{x^2}{4} - \frac{y^2}{9/4} = 1 \Rightarrow \frac{x^2}{4} - \frac{4y^2}{9} = 1 \Rightarrow 9x^2 - 16y^2 = 36 \).

(vi) distances between the foci and directrices are \( 4\sqrt{7} \) and \( \frac{16}{\sqrt{7}} \) respectively.

Solution

Distance between foci: \( 2ae = 4\sqrt{7} \quad \rightarrow (1) \)

Distance between directrices: \( \frac{2a}{e} = \frac{16}{\sqrt{7}} \quad \rightarrow (2) \)

Multiply (1) and (2): \( 4a^2 = 4\sqrt{7} \times \frac{16}{\sqrt{7}} = 64 \Rightarrow a^2 = 16 \).

From (1): \( a^2 e^2 = (2\sqrt{7})^2 = 28 \Rightarrow a^2 + b^2 = 28 \Rightarrow 16 + b^2 = 28 \Rightarrow b^2 = 12 \).

Equation: \( \frac{x^2}{16} - \frac{y^2}{12} = 1 \Rightarrow 3x^2 - 4y^2 = 48 \).

(vii) eccentricity is \( \frac{\sqrt{13}}{3} \) and the sum of squares of the lengths of axes is 52.

Solution

\( (2a)^2 + (2b)^2 = 52 \Rightarrow 4a^2 + 4b^2 = 52 \Rightarrow a^2 + b^2 = 13 \quad \rightarrow (1) \)

\( e = \frac{\sqrt{13}}{3} \Rightarrow e^2 = \frac{13}{9} \Rightarrow 1 + \frac{b^2}{a^2} = \frac{13}{9} \Rightarrow \frac{a^2 + b^2}{a^2} = \frac{13}{9} \).

Substitute (1): \( \frac{13}{a^2} = \frac{13}{9} \Rightarrow a^2 = 9 \).

From (1): \( 9 + b^2 = 13 \Rightarrow b^2 = 4 \).

Equation: \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \Rightarrow 4x^2 - 9y^2 = 36 \).

(viii) transverse axis is \( 2a \) and the vertex bisects the line segment joining the centre and focus.

Solution

Vertex bisects centre and focus \( \Rightarrow a = \frac{ae}{2} \Rightarrow e = 2 \).

\( e^2 = 1 + \frac{b^2}{a^2} \Rightarrow 4 = 1 + \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = 3 \Rightarrow b^2 = 3a^2 \).

Equation: \( \frac{x^2}{a^2} - \frac{y^2}{3a^2} = 1 \Rightarrow 3x^2 - y^2 = 3a^2 \).

(ix) which passes through the point \( (2,1) \) and whose eccentricity is \( \sqrt{\frac{3}{2}} \).

Solution

Passes through \( (2,1) \): \( \frac{4}{a^2} - \frac{1}{b^2} = 1 \Rightarrow 4 - \frac{a^2}{b^2} = a^2 \Rightarrow \frac{a^2}{b^2} = 4 - a^2 \quad \rightarrow (1) \)

\( e = \sqrt{\frac{3}{2}} \Rightarrow e^2 = \frac{3}{2} \Rightarrow 1 + \frac{b^2}{a^2} = \frac{3}{2} \Rightarrow \frac{b^2}{a^2} = \frac{1}{2} \quad \rightarrow (2) \)

Multiply (1) and (2): \( 1 = \frac{1}{2}(4 - a^2) \Rightarrow 2 = 4 - a^2 \Rightarrow a^2 = 2 \).

From (2): \( \frac{b^2}{2} = \frac{1}{2} \Rightarrow b^2 = 1 \).

Equation: \( \frac{x^2}{2} - \frac{y^2}{1} = 1 \Rightarrow x^2 - 2y^2 = 2 \).

(x) eccentricity is 3 and the co-ordinates of one focus are \( \left(\frac{3}{2}, 0\right) \).

Solution

\( ae = \frac{3}{2} \Rightarrow a \times 3 = \frac{3}{2} \Rightarrow a = \frac{1}{2} \).

\( e = 3 \Rightarrow e^2 = 9 \Rightarrow 1 + \frac{b^2}{a^2} = 9 \Rightarrow \frac{b^2}{a^2} = 8 \Rightarrow b^2 = 8a^2 = 8 \times \frac{1}{4} = 2 \).

Equation: \( \frac{x^2}{1/4} - \frac{y^2}{2} = 1 \Rightarrow 4x^2 - \frac{y^2}{2} = 1 \Rightarrow 8x^2 - y^2 = 2 \).

(xi) eccentricity is \( \sqrt{\frac{3}{2}} \) and the length of semi latus rectum is 2.

Solution

Semi-latus rectum \( \frac{b^2}{a} = 2 \Rightarrow b^2 = 2a \quad \rightarrow (1) \)

\( e = \sqrt{\frac{3}{2}} \Rightarrow e^2 = \frac{3}{2} \Rightarrow 1 + \frac{b^2}{a^2} = \frac{3}{2} \Rightarrow \frac{b^2}{a^2} = \frac{1}{2} \).

Substitute (1): \( \frac{2a}{a^2} = \frac{1}{2} \Rightarrow \frac{2}{a} = \frac{1}{2} \Rightarrow a = 4 \Rightarrow a^2 = 16 \).

\( b^2 = 2 \times 4 = 8 \).

Equation: \( \frac{x^2}{16} - \frac{y^2}{8} = 1 \Rightarrow x^2 - 2y^2 = 16 \).

(xii) co-ordinates of foci are \( (5,0), (-5,0) \) and the eccentricity is \( \frac{5}{4} \).

Solution

\( ae = 5 \Rightarrow a^2 e^2 = 25 \Rightarrow a^2\left(1 + \frac{b^2}{a^2}\right) = 25 \Rightarrow a^2 + b^2 = 25 \quad \rightarrow (1) \)

\( e = \frac{5}{4} \Rightarrow a^2 \times \left(\frac{5}{4}\right)^2 = 25 \Rightarrow a^2 \times \frac{25}{16} = 25 \Rightarrow a^2 = 16 \quad \rightarrow (2) \)

From (1) and (2): \( 16 + b^2 = 25 \Rightarrow b^2 = 9 \).

Equation: \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \Rightarrow 9x^2 - 16y^2 = 144 \).

© S N Dey Mathematics Class 11 Solutions | Hyperbola