Question

Find the equation of the hyperbola, the length of whose latustrectum is 8 and eccentricity is $3/\sqrt{5}$.Also determine the equation of directrices.

Or

Find the equation of the ellipse whose axes are along the coordinate axes,vertices are $\pm 5,0)$and foci at $(\pm 4,0)$. Also determine the length of major and minor axes.

Answer 1

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

Then, length of latusrectum =8

$\Rightarrow \frac{2b^2}{a}=8\Rightarrow b^2=4a$

$\Rightarrow a^2(e^2-1)=4a$

$\Rightarrow a^2(\frac{9}{5}-1)=4a$

$\Rightarrow \frac{4a^2}{5}=4a[\because e=3/\sqrt{5}]$

$\Rightarrow a=5$

On substituting a = 5 in $b^2=4a$, we get $b^2=20$.

Hence, the equation of the required hyperbola is

$\frac{x^2}{25}-\frac{y^2}{20}=1$

$\therefore$ The equation of directrices are x=$\pm \frac{a}{e},$

$\therefore x=\pm \frac{5}{\frac{3}{\sqrt{5}}}\Rightarrow x=\pm \frac{5\sqrt{5}}{3}$

Or

Let the equation of the required ellipse be

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

The coordinates of its vertices and foci are $(\pm a,0)$ adn $(\pm c,0)$, respectively.

According to the question,

$\therefore a=5andc=4$

Now, $c^2=a^2-b^2\Rightarrow b^2=a^2-c^2$

$\Rightarrow b^2=(5{)}^2-(4{)}^2$

$\Rightarrow b^2=25-16=9$

Substituting the value of $a^2$and $b^2$in Eq.(i) we get

$\frac{x^2}{25}+\frac{y^2}{9}=1$, which is the quation of the required ellipse.

Here, we find that a>b.

$\therefore$The length of major axis $2a=2\times 5=10$

and length of minor axis $2b=2\times 3=6$