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If P is a point on the parabola \mathrm{y^2=4 a x} such that the lengths of the subtangent and the subnormals at P are equal, then point P is

Option: 1

\mathrm{(a, 2 a) \text { or }(a,-2 a)}


Option: 2

\mathrm{(4 a,-4 a) \text { or }(-4 a, 4 a)}


Option: 3

\mathrm{(2 a, \sqrt{2} a) \text { or }(2 a,-\sqrt{2} a)}


Option: 4

\mathrm{\text { none of these }}


Answers (1)

best_answer

Let \mathrm{P\left(a t^2, 2 a t\right)} be a point on \mathrm{y^2=4 a x.}

The equation of tangent at P is

\mathrm{ y t=x+a t^2 }                                 .........(1)

The equation of normal at P is

\mathrm{ y-2 a t=-t\left(x-a t^2\right) }         .........(2)

\mathrm{\therefore \quad}  the coordinates of T are

\mathrm{ y=0, x=-a t^2 \quad \Rightarrow T\left(-a t^2, 0\right) }

The coordinates of N are

\mathrm{ y=0, x=a t^2+2 a \Rightarrow N\left(a t^2+2 a, 0\right) }

M is \mathrm{\left(a t^2, 2 a t\right)}

\mathrm{\therefore \quad \, \, length\, \, of \, \, subtangent =M T=a t^2+a t^2=2 a t^2}

And, length of subnormal \mathrm{=M N=a t^2+2 a-a t^2=2 a}

As given \mathrm{2 a t^2=2 a \Rightarrow a t^2=a ; t= \pm 1}

\mathrm{\therefore \quad P} is \mathrm{\left(a t^2, 2 a t\right)=(a, 2 a),(a,-2 a).}

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Sanket Gandhi

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