VITEEE 2017 Paper

The line

x

p

+

y

q

= 1 will be a normal to the parabola y2 = 4ax if, for some value of m, it is identical with
y = mx - 2am - am3 i.e. mx - y = (2am + am3)
Comparing coefficients, we get

m

1∕p

=

−1

1∕q

=

2a,m+am3

1

⇒ mp = - q
∴ m = −

q

p

and mp = m (2a + am2)
or P = 2a + am2 = 2a + a (−

q

p

)2
or p = 2a +

aq2

p2

or p3 = 2ap2+aq2
Which is the required condition