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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


6 years ago

IIT Delhi
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 Qslope(PQ) = m and the slope of the tangent line at Q is -1/nx = the angle that PQ makes with the horizontaly = the angle that the tangent at Q makes with the horizontalx-y = the angle we are looking forWith tan (x) = m and tan(y) = -1/n, use the identity$\tan (x-y)=\frac{\tan x- \tan y}{1+\tan x *\tan y}$

6 years ago
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions