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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

Varun , 11 Years ago
Grade 12
anser 1 Answers
Sher Mohammad

Last Activity: 11 Years ago

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}

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