Question

If a directrix of a hyperbola centered at the origin passing through the point $\left( 4,-2\sqrt{3} \right)$ is $5x=4\sqrt{5}$ and its eccentricity is e, then:
(a) $4{{e}^{4}}-24{{e}^{2}}+35=0$,
(b) $4{{e}^{4}}+8{{e}^{2}}-35=0$,
(c) $4{{e}^{4}}-12{{e}^{2}}-27=0$,
(d) $4{{e}^{4}}-24{{e}^{2}}-27=0$.

Answer 1

Hint: We start solving the problem by recalling the definitions of hyperbola, eccentricity and directrix with center at origin. We compare the standard form of directrix and the equation of given directrix to get the value of a in terms of eccentricity. We then use eccentricity to find the value of b in terms of eccentricity. We then substitute the obtained values of a, b and the given point in the equation of hyperbola and make necessary calculations to get the required result.

Complete step-by-step answer:
According to the problem, we have a hyperbola centered at origin and passing through the point $\left( 4,-2\sqrt{3} \right)$. The equation of directrix of this hyperbola is $5x=4\sqrt{5}$ and eccentricity is e.
Let us draw all the given information.

We know that the equation of the hyperbola centered at origin is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and we know that the eccentricity of the hyperbola is $e=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}$ ---(1).
We know that equation of one of the directrix of the hyperbola is given as $x=\dfrac{a}{e}$ ---(2).
According to the problem, we have the equation of directrix of this hyperbola is $5x=4\sqrt{5}$.
$\Rightarrow x=\dfrac{4\sqrt{5}}{5}$.
$\Rightarrow x=\dfrac{4}{\sqrt{5}}$ ---(3).
Comparing equations (2) and (3), we get $\dfrac{a}{e}=\dfrac{4}{\sqrt{5}}$.
$a=\dfrac{4e}{\sqrt{5}}$.
${{a}^{2}}=\dfrac{16{{e}^{2}}}{5}$ ---(4).
From equation (1), we have $e=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}$.
$\Rightarrow {{e}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}$.
$\Rightarrow {{a}^{2}}{{e}^{2}}={{a}^{2}}+{{b}^{2}}$.
$\Rightarrow {{a}^{2}}{{e}^{2}}-{{a}^{2}}={{b}^{2}}$.
$\Rightarrow {{a}^{2}}\left( {{e}^{2}}-1 \right)={{b}^{2}}$.
From equation (4),
$\Rightarrow \dfrac{16{{e}^{2}}\left( {{e}^{2}}-1 \right)}{5}={{b}^{2}}$ ---(5).
We substitute equations (4) and (5) in the equation of hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.
$\Rightarrow \dfrac{{{x}^{2}}}{\dfrac{16{{e}^{2}}}{5}}-\dfrac{{{y}^{2}}}{\dfrac{16{{e}^{2}}\left( {{e}^{2}}-1 \right)}{5}}=1$.
$\Rightarrow \dfrac{{{x}^{2}}}{1}-\dfrac{{{y}^{2}}}{\left( {{e}^{2}}-1 \right)}=\dfrac{16{{e}^{2}}}{5}$ ---(6).
We substitute the point $\left( 4,-2\sqrt{3} \right)$ in equation (6), as the equation of hyperbola passes through it.
$\Rightarrow \dfrac{{{\left( 4 \right)}^{2}}}{1}-\dfrac{{{\left( -2\sqrt{3} \right)}^{2}}}{\left( {{e}^{2}}-1 \right)}=\dfrac{16{{e}^{2}}}{5}$.
$\Rightarrow 16-\dfrac{12}{\left( {{e}^{2}}-1 \right)}=\dfrac{16{{e}^{2}}}{5}$.
\[\Rightarrow \dfrac{4\left( {{e}^{2}}-1 \right)-3}{\left( {{e}^{2}}-1 \right)}=\dfrac{4{{e}^{2}}}{5}\]
\[\Rightarrow 5\times \left( 4{{e}^{2}}-4-3 \right)=4{{e}^{2}}\times \left( {{e}^{2}}-1 \right)\].
\[\Rightarrow 5\times \left( 4{{e}^{2}}-7 \right)=4{{e}^{2}}\times \left( {{e}^{2}}-1 \right)\].
\[\Rightarrow 20{{e}^{2}}-35=4{{e}^{4}}-4{{e}^{2}}\].
\[\Rightarrow 4{{e}^{4}}-4{{e}^{2}}-20{{e}^{2}}+35=0\].
\[\Rightarrow 4{{e}^{4}}-24{{e}^{2}}+35=0\].
We have found the condition for eccentricity of hyperbola as \[4{{e}^{4}}-24{{e}^{2}}+35=0\].
∴ The condition for eccentricity of hyperbola is \[4{{e}^{4}}-24{{e}^{2}}+35=0\].

So, the correct answer is “Option A”.

Note: We can also solve the problem by finding the value of eccentricity using the value of $\dfrac{a}{e}$ and substituting it in the options to verify which is the correct option. We need to make sure about the center of the given hyperbola. If the center changes, then the total answer changes with a lot of deviation from the answer we just had. Similarly, we expect to find the values of a, b and also the equation of the axes, foci of the given hyperbola.