Question

The foci of a hyperbola coincide with foci of the ellipse $\frac{x^{2}}{25}$+ $\frac{y^{2}}{4}$= 1. If its eccentricity is 2 then its equation must be-

• A. $\frac{x^{2}}{4}$– $\frac{y^{2}}{12}$= 1
• B. $\frac{x^{2}}{12}$–$\frac{y^{2}}{4}$= 1
• C. $\frac{x^{2}}{4}$–$\frac{y^{2}}{12}$= –1
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

$\frac{x^{2}}{25}$+$\frac{y^{2}}{4}$= 1 … (1)

a₁² = 25 ; b₁² = 4

e₁ = $\sqrt{1 - \frac{b_{1}^{2}}{a_{1}^{2}}}$= $\sqrt{1 - \frac{4}{25}}$= $\frac{\sqrt{21}}{5}$

\ Focus of ellipse (1) Ž S (± ae, 0)

Ž 5 $\left\{ \pm 5\frac{\sqrt{21}}{5},0 \right\}$= (±$\sqrt{21}$, 0)

\ Hyperbola is Ž $\frac{x^{2}}{{a_{2}}^{2}}$ – $\frac{y^{2}}{{b_{2}}^{2}}$= 1 … (2)

Given a₁e₁ = a₂e₂ \ e₂ = 2

Ž $\sqrt{21}$= 2a₂ Ž $a_{2}^{2}$= $\frac{21}{4}$

Ž b₂² = a₂² (e₂² –1) = $\frac{21}{4}$ (4 –1) = $\frac{21}{4}$× 3 = $\frac{63}{4}$

\ equation of hyperbola is

Ž $\frac{x^{2}}{(21/4)}$–$\frac{y^{2}}{(63/4)}$= 1 Ž $\frac{x^{2}}{21}$– $\frac{y^{2}}{63}$= $\frac{1}{4}$