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Equation of Tangent at a Point (x,y) in Terms of f'(x)

If the tangen...

Question

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

t = - c o s e c θ

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t = - sec θ

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t = 2 tan θ

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t = 2 cot θ

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Solution

The correct options are
B $t = - c o s e c θ$
D $t = 2 tan θ$
We know equation of tangent is $y = \frac{x}{t} + a t$ and normal is $x c o s θ + y c o t θ = 2 a$
So, $(a t^{2}, 2 a t)$ must satisfy the equation of normal
$\Rightarrow t^{2} c o s θ + 2 t \times c o t θ - 2 = 0$
So, the roots are $t = - c o s e c θ$ and $t = 2 t a n θ$

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