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  • If normal at the point <img alt="\mathrm{P\left(a t^2, 2 a t\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BP%5Cleft%28a%20t%5E2%2C%202%20a%20t

If normal at the point \mathrm{P\left(a t^2, 2 a t\right)} to the parabola \mathrm{y^2=4 a x} cuts the circle drawn with PS as diameter at Q (where S is the focus of the parabola. Then \mathrm{P Q=k^a \sqrt{1+t^2}}, where k =

 

Option: 1

1


Option: 2

2


Option: 3

\mathrm{1/2}


Option: 4

3


Answers (1)

best_answer

If M is the perpendicular from S to tangent at P, then PQSM is a square and SQ = SM. 

Tangent at the point P will be\mathrm{t y=x+a t^2}

length of perpendicular from \mathrm{S \text { to } \mathrm{PM}=\frac{a+a t^2}{\sqrt{1+t^2}}=a \sqrt{1+t^2}}

 

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