Question

If the tangent to $y^2=4ax$ at the point $(a{t}^2,2at)$ where $|t|>1$ is a normal to $x^2-y^2=a^2$ at the point $(a\sec \theta ,a\tan \theta ),$ then

• A. $t=-\text{cosec}\theta$
• B. $t=-\sec \theta$
• C. $t=2\tan \theta$
• D. $t=2\cot \theta$

Answer 1

Answer: (A), (C)

The correct options are
A $t=-\text{cosec}\theta$
C $t=2\tan \theta$

Equation of tangent $y^2=4ax$ at the point $(a{t}^2,2at)$ is $2aty=2a(x+a{t}^2)$
$\Rightarrow ty=x+a{t}^2$
$\Rightarrow y=\frac{x}{t}+at\cdots \left(1\right)$

Equation of normal to $x^2-y^2=a^2$ :
$\frac{dx}{dy}=\frac{y}{x}$
${[-\frac{dx}{dy}]}_{(a\sec \theta ,a\tan \theta )}=-\frac{a\tan \theta }{a\sec \theta }=-\sin \theta$

Therefore, equation of the normal is
$y-a\tan \theta =-\sin \theta (x-a\sec \theta )$
$\Rightarrow y=-\frac{x}{\text{cosec}\theta }+2a\tan \theta \cdots \left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$, we get
$t=-\text{cosec}\theta$ and $t=2\tan \theta$