Question

The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity $\frac{5}{4}$ is

• A. $\frac{(x - 1)^{2}}{16} - \frac{(y - 5)^{2}}{9}$= 1
• B. $\frac{x^{2}}{16} - \frac{y^{2}}{9}$ = 1
• C. $\frac{(x - 1)^{2}}{16} - \frac{(y - 5)^{2}}{9}$ = –1
• D. None of these

Answer 1

Answer: (A)

$\frac{(x - 1)^{2}}{16} - \frac{(y - 5)^{2}}{9}$= 1

Answer 2

Here S₁ ≡ ( 6, 5); S₂ ≡ (–4, 5) , e = 5/ 4, so that S₁S₂ = 10 ⇒ 2ae = 10 ⇒ a = 4.

And b² = a²(e² – 1) = 16$\left( \frac{25}{16} - 1 \right)$ = 9.

Centre of the curve is (1, 5) ⇒ Equation of required hyperbola is;

$\frac{(x - 1)^{2}}{16} - \frac{(y - 5)^{2}}{9} = 1$.