Question

Find the locus of intersection of normals to a parabola inclined at
complementary angles to the axis.​

Answer 1

Answer:

The given parabola is y2=4ax

The equation of any normal to the above parabola is

y=mx−2am−am3

If P(x1,y1) is a point on the locus then

y1=mx1−2am−am3

Or

am3+(2a−x1)m+y1=0(1)

This is a cubic equation in m. Let its roots be m1,m2 and m3 which are the slopes of the normals to the given parabola through P . We have

m1m2m3=−y1a(2)

If two of the normals, say with slopes m1 and m2 makes complementary angles θ and (90o−θ) with the axis of the parabola then

m1=tanθ,m2=tan(90o−θ)=cotθ

m1m2=tanθcotθ=1(3)

Dividing equation (2) by (3), we get

m3=−y1a

But m3 is the root of (2). Hence

y21=a(x1−a)

The equation of the locus of P is therefore

y2=a(x−a)

Step-by-step explanation:

ihope this helps you