Question

A hyperbola has center at origin and passing through $(4,-2\sqrt{3})$ and having directrix $5x=4\sqrt{5}$ then eccentricity of hyperbola (e) satisfy the equation

• A. $4e^4-24e^2+35=0$
• B. $4e^4+24e^2-35=0$
• C. $4e^4-24e^2-35=0$
• D. $4e^4+24e^2+35=0$

Answer 1

Answer: (A)

$4e^4-24e^2+35=0$

Given directrix is $5x=4\sqrt{5}⟹x=\frac{4}{\sqrt{5}}⟹\frac{a}{e}=\frac{4}{\sqrt{5}}$

$⟹a=\frac{4e}{\sqrt{5}}$

$b^2=a^2(e^2-1)=\frac{16e^2(e^2-1)}{5}$

Let the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$(4,-2\sqrt{3})$ lies on hyperbola

$⟹\frac{16}{a^2}-\frac{12}{b^2}=1$

$⟹\frac{5}{e^2}-\frac{15}{4e^2(e^2-1)}=1⟹\frac{5}{e^2}(\frac{4e^2-7}{4(e^2-1)})=1$

$⟹4e^4-4e^2=20e^2-35⟹4e^4-24e^2+35=0$