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Grade: 12
        I am not able to solve the following problem

#1) Prove that the normal to parabola y^2=4ax at (am^2,-2am) intersects the parabola again at an angle tan-1(m/2)

What I am thinking is to solve the equation of parabola and equation of normal y=mx-am^-3 -2am simultaneously and at that point I will find the slope of tangent and will get the angle between tangent and normal. The problem is that answer is not coming.

#2) For what values of a will the tangents drawn to parabola y^2=4ax from a point , not on the y-axis, will be normals to the parabola x^2=4y?

I have no idea on how to solve this question
Thanks in Advance
5 years ago

Answers : (1)

Sher Mohammad
IIT Delhi
askIITians Faculty
174 Points
							Let     P\left(a m^2,-2 a m\right) , Q\left(a n^2,-2 a n\right)
The normal at P intersects the parabola again at Q
slope(PQ) = m and the slope of the tangent line at Q is -1/n
x = the angle that PQ makes with the horizontal
y = the angle that the tangent at Q makes with the horizontal
x-y = the angle we are looking for
With tan (x) = m and tan(y) = -1/n, use the identity
\tan (x-y)=\frac{\tan  x- \tan  y}{1+\tan  x *\tan  y}
5 years ago
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