Question

For a parabola y^2=4ax two variable chords pq and rs at right angles are drawn through the fixed

Answer 1

They are drawn through the fixed directrix

Answer 2

Answer:

For a given parabola

y2=4ax, two variable chords PQ and RS at right angles are drawn through the fixed point A(x1, y1) inside the parabola, making variable angles θ and α with x-axis. If r1, r2, r3, r4 are distances of P, Q, R and S from A, then the value of

1/r1r2+1/r3r4

Let two perpendicular chords through (x1, y1) be PQ and RS

Equation of PQ is x−x1/cos θ=y−y1/sin θ=r

Where tan θ = slope of PQ

Any point on this line may be taken as

(x1+r cos θ, y1+r sin θ)

As the point lies on y2=4ax,

(y1+r sin θ)2=4a(x1+rcos θ)

r2 sin2 θ+2y1 sin θr−4a cos θr+y21−4ax1=0

r1r2=(y21−4ax1)/sin2 θ

Similarly

r3r4=(y21−4ax1)/sin2α where tan α = slope of RS

Since PQ is perpendicular to RS,

α=90+θ or θ−90

In either case; sin2α=cos2θ

Now

1/r1r2+1/r3r4=sin2θ+cos2θ/(y21−4ax1)=1/(y21−4ax1),

1/r1r2+1/r3r4=sin2θ+cos2θ/(y21−4ax1)=1/(y21−4ax1), which is constant.