Question

Equation of the hyperbola passing through $(2,1)$ and having distance between the directrices $\frac{4}{\sqrt{3}}$ is

Answer 1

We have

Point are $\Rightarrow (1,2)$

Distance between directrix $\Rightarrow \frac{4}{\sqrt{3}}$

We know that,

Equation of Hyperbola is

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

As it passes through $(2,1)$

Then,

$\frac{4}{a^2}-\frac{1}{b^2}=1$

$b^2=\frac{a^2}{4-a^2}......\left(1\right)$

Distance between directrix $=\frac{4\sqrt{3}}{3}$

$\Rightarrow 2ae=\frac{4\sqrt{3}}{3}$

$\Rightarrow a^2{e}^2\frac{4}{3}......\left(2\right)$

Also,

$e^2=\frac{a^2+b^2}{a^2}$

$\Rightarrow a^2+b^2=a^2{e}^2$

$\Rightarrow a^2+b^2=\frac{4}{3}$ using equation (2)

Now,

$a^2+\frac{a^2}{4-a^2}=\frac{4}{3}$

$\Rightarrow \frac{4a^2-a^4+a^3}{4-a^2}=\frac{4}{3}using\left(1\right)$

$\Rightarrow \frac{5a^2-a^4}{4-a^2}=\frac{4}{3}$

$\Rightarrow 15a^2-3a^4=16-4a^2$

$\Rightarrow 3a^4-19a^2+16=0$

$\Rightarrow 3a^4-3a^2-16a^2+16=0$

$\Rightarrow 3a^2(a^2-1)-16(a^2-1)=0$

$\Rightarrow (a^2-1)(3a^2-1)=0$

$\Rightarrow a^2-1=0,3a^2-1=0$

$\Rightarrow a^2=1,a^2=\frac{1}{3}$

When put

$a^2=1$

$b^2=\frac{1}{4-1}=\frac{1}{3}$

So,

Equation of Hyperbola

$\frac{x^2}{\frac{1}{1}}-\frac{y^2}{\frac{1}{3}}=1$

$\Rightarrow x^2-3y^2=1$

Hence, this is the answer.