The Equation of the Hyperbola Whose Foci Are (6, 4) and (−4, 4) and Eccentricity 2, is - Mathematics

MCQ

The equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity 2, is

Options

• $\frac{(x - 1 )^2}{25/4} - \frac{(y - 4 )^2}{75/4} = 1$

• $\frac{(x + 1 )^2}{25/4} - \frac{(y + 4 )^2}{75/4} = 1$

• $\frac{(x - 1 )^2}{75/4} - \frac{(y - 4 )^2}{25/4} = 1$

• none of these

Solution

$\frac{(x - 1 )^2}{25/4} - \frac{(y - 4 )^2}{75/4} = 1$

The centre of the hyperbola is the midpoint of the line joining the two foci.
So, the coordinates of the centre are $\left( \frac{6 - 4}{2}, \frac{4 + 4}{2} \right), i . e . \left( 1, 4 \right) .$

Let 2a and 2b be the length of the transverse and the conjugate axes, respectively. Also, let e be the eccentricity.

$\Rightarrow \frac{\left( x - 1 \right)^2}{a^2} - \frac{\left( y - 4 \right)^2}{b^2} = 1$

Now, distance between the two foci = 2ae

$2ae = \sqrt{\left( 6 + 4 \right)^2 + \left( 4 - 4 \right)^2}$

$\Rightarrow 2ae = 10$

$\Rightarrow ae = 5$

$\Rightarrow a = \frac{5}{2}$

$\text { Also }, b^2 = \left( ae \right)^2 - \left( a \right)^2$

$\Rightarrow b^2 = 25 - \left( \frac{25}{4} \right)$

$\Rightarrow b^2 = \frac{75}{4}$

Equation of the hyperbola is given below:

$\frac{\left( x - 1 \right)^2}{25/4} - \frac{\left( y - 4 \right)^2}{75/4} = 1$

Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Q 14 | Page 19