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Find the equation of the tangent and normal to the parabola `y^2=4a x` at the point `(a t^2,\ 2a t)` .

Text Solution

Verified by Experts

The correct Answer is:
`y=-t x+2 a t+a t^{3}`
Given, equation of parabola `y^{2}=4 ax`
Differentiating w.r.t. `x`, $$ \begin{aligned} &2 y \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{a} \\ &\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{y} \end{aligned} $$ ...
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Knowledge Check

  • The equation of normal to the parabola y^2=4ax at the point (at^2, 2at) is

    A
    `y+2at^2=xt-at^2`
    B
    `y-2at=xt-at^2`
    C
    `y-2at= -xt+at^3`
    D
    None of the above

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