Question

A normal to the hyperbola $x^2-4y^2=4$ meets the $x$ and $y$ axes at $A$ and $B$ respectively. The locus of point of intersection of the straight lines drawn through $A$ and $B,$ perpendicular to the $x$ and $y$ axes respectively is a hyperbola with

• A. eccentricity equal to $\sqrt{3}$
• B. length of latus rectum equal to $20$
• C. equation of auxiliary circle equal to $x^2+y^2=25$
• D. distance between foci equal to $5\sqrt{5}$

Answer 1

Answer: (B), (D)

distance between foci equal to $5\sqrt{5}$

The hyperbola is $\frac{x^2}{4}-\frac{y^2}{1}=1$
$a=2,b=1$
So, equation of normal at any point $P\left(\theta \right)$ is
$2x\cos \theta +y\cot \theta =2^2+1^2=5$
So, $A(\frac{5}{2}\sec \theta ,0)$ and $B(0,5\tan \theta )$
$\Rightarrow$ Equation of lines drawn from $A$ and $B$ perpendicular to $x-$axis and $y-$axis respectively, are
$x=\frac{5}{2}\sec \theta ...\left(1\right)$
$y=5\tan \theta ...\left(2\right)$

Using ${\sec }^2\theta -{\tan }^2\theta =1$ and eqns. $\left(1\right)$ and $\left(2\right),$ we get
$\frac{4x^2}{25}-\frac{y^2}{25}=1$
or $\frac{x^2}{25/4}-\frac{y^2}{25}=1$, which is equation of hyperbola.

Now, $e_h^2=1+\frac{b^2}{a^2}=1+\frac{25}{25/4}=5$
$\Rightarrow e_h=\sqrt{5}$

Length of latus rectum $=\frac{2b^2}{a}=\frac{2\times 25}{5/2}=4\times 5=20$

Equation of auxiliary circle of hyperbola is $x^2+y^2=\frac{25}{4}$

Distance between foci $=2a{e}_h=2\left(\frac{5}{2}\right)\sqrt{5}=5\sqrt{5}$