A normal to the hyperbola x2-4y2=4 meets the x and y axes at A and B. The locus of the point of intersection of the straight lines drawn through A and B perpendicular to the x and y axes respectively is

Dear Student,



Equation of hyperbola is $x^2-4y^2=4$
$$\begin{gathered} \Rightarrow \frac{x^2}{4}-\frac{4y^2}{4}=\frac{4}{4} \\ \Rightarrow \frac{x^2}{4}-\frac{y^2}{1}=1 \\ \Rightarrow \frac{x^2}{2^2}-\frac{y^2}{1^2}=1 \end{gathered}$$
The equation of normal at the point $Q\left(2sec\left(\theta \right), \tan \left(\theta \right)\right)$ is given as $\frac{y}{\tan \left(\theta \right)}+\frac{4x}{2sec\left(\theta \right)}=4+1\Rightarrow \frac{y}{\tan \left(\theta \right)}+\frac{2x}{sec\left(\theta \right)}=5$.
At, $y=0, \frac{y}{\tan \left(\theta \right)}+\frac{2x}{sec\left(\theta \right)}=5\Rightarrow x=\frac{5sec\left(\theta \right)}{2}$
At, $x=0, \frac{y}{\tan \left(\theta \right)}+\frac{2x}{sec\left(\theta \right)}=5\Rightarrow y=5\tan \left(\theta \right)$
The normal meets the x-axis at $A\left(\frac{5sec\left(\theta \right)}{2}, 0\right)$ and y-axis at $B\left(0, 5\tan \left(\theta \right)\right)$.
Equation of the line through A and perpendicular to x-axis, AP is given as $x=\frac{5}{2}sec\left(\theta \right)\Rightarrow sec\left(\theta \right)=\frac{2x}{5} ....\left(i\right)$
Equation of the line through B ​and perpendicular to y-axis, BP is given as $y=5\tan \left(\theta \right)\Rightarrow \tan \left(\theta \right)=\frac{y}{5} ....\left(ii\right)$
Squaring and subtracting equation (i) and (ii) we get

$$\begin{gathered} se{c}^2\left(\theta \right)-{\tan }^2\left(\theta \right)=1 \\ \Rightarrow {\left(\frac{2x}{5}\right)}^2-{\left(\frac{y}{5}\right)}^2=1 \\ \Rightarrow \frac{4x^2}{25}-\frac{y^2}{25}=1 \\ \Rightarrow 4x^2-y^2=25 \end{gathered}$$
The required locus of P is $4x^2-y^2=25$
Regards,