# For what values of 'a' will the tangents drawn to the parabola y2 = 4ax from a point, not on the y axis, will be normal to the parabola x2=4y.

Vikas TU
14149 Points
7 years ago
tangent for y^2 =4ax=>
2ydy/dx = 4a
or
dy/dx = 2a/y  = m (let)
and tangent for x^2 = 4y
2x = 4dy/dx
dy/dx = x/2 = n (let)
Now given,
m * n = -1
we get,
y = – ax put it in y^2 = 4ax eqn. we get
a^2x^2 – 4ax = 0
ax(ax – 4 ) = 0
x  = 0  and x = a/4
y= 0 and y = a
Now put x and y values in x^2 = 4y then,
a^2/16 = 4a
a^2 – 64a = 0
a(a – 64) = 0
a= 0 and a= 64.
shuaib
22 Points
6 years ago
but the answer is  a2√2.................... ( except in the y axis)...
i think m*n = -1 … is wrong
shuaib
22 Points
6 years ago
let the equation of the parabola1
P1 : y2= 4ax..                    tangent equation.==>  y=mx + a/m..............................................(1)
P2 : X= 4y..................normal equation..==>y = mx + 2a +a/m2
= mx +2 + 1/m2        (here a=1)..................(2)
from (1) and (2) ….. we can conclude
mx + a/m = mx + 2 + 1/m2
…........on solving we get a quadratic equation ….....
2m- am +1 =0       …..............
so m should be a real and distinct number if the D >0...
so that …...... (-a)2 -4(2)(1) > 0.
which gives (a – 2root 2)(a + 2root 2) >0
…........
whch finally tells  a is an elment of (-infinity, -2root 2) or (2root 2 , infinity).....answer