Hyperbola 1 Question 2
3. If a directrix of a hyperbola centred at the origin and passing through the point $(4,-2 \sqrt{3})$ is $5 x=4 \sqrt{5}$ and its eccentricity is $e$, then
(2019 Main, 10 April I)
(a) $4 e^{4}-12 e^{2}-27=0$
(b) $4 e^{4}-24 e^{2}+27=0$
(c) $4 e^{4}+8 e^{2}-35=0$
(d) $4 e^{4}-24 e^{2}+35=0$
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Answer:
Correct Answer: 3. (a)
Solution:
- Let the equation of hyperbola is
$$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $$
Since, equation of given directrix is $5 x=4 \sqrt{5}$
$$ \begin{aligned} \text { so } & 5 \frac{a}{e} & =4 \sqrt{5} \quad\left[\because \text { equation of directrix is } x=\frac{a}{e}\right] \\ \Rightarrow & \frac{a}{e} & =\frac{4}{\sqrt{5}} \end{aligned} $$
and hyperbola (i) passes through point $(4,-2 \sqrt{3})$
$$ \text { so, } \quad \frac{16}{a^{2}}-\frac{12}{b^{2}}=1 $$
The eccentricity $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$$ \begin{array}{lrl} \Rightarrow & e^{2}=1+\frac{b^{2}}{a^{2}} \\ \Rightarrow & a^{2} e^{2}-a^{2}=b^{2} \end{array} $$
From Eqs. (ii) and (iv), we get
$$ \frac{16}{5} e^{4}-\frac{16}{5} e^{2}=b^{2} $$
From Eqs. (ii) and (iii), we get
$$ \begin{aligned} & \frac{16}{\frac{16}{5} e^{2}}-\frac{12}{b^{2}}=1 \Rightarrow \frac{5}{e^{2}}-\frac{12}{b^{2}}=1 \\ \Rightarrow \quad & \frac{12}{b^{2}}=\frac{5}{e^{2}}-1 \Rightarrow \frac{12}{b^{2}}=\frac{5-e^{2}}{e^{2}} \\ \Rightarrow \quad & b^{2}=\frac{12 e^{2}}{5-e^{2}} \end{aligned} $$
From Eqs. (v) and (vi), we get
$$ \begin{aligned} & 16 e^{4}-16 e^{2}=5 \frac{12 e^{2}}{5-e^{2}} \Rightarrow 16\left(e^{2}-1\right)\left(5-e^{2}\right)=60 \\ \Rightarrow & 4\left(5 e^{2}-e^{4}-5+e^{2}\right)=15 \\ \Rightarrow & 4 e^{4}-24 e^{2}+35=0 \end{aligned} $$
