Question

If two distinct chords of a parabola $y^2=4ax$, passing through (a, 2a) are bisected on the line $x+y=1$, then length of the latus-rectum can be

• A. 2
• B. 1
• C. 4
• D. 3

Answer 1

Answer: (A), (B), (D)

The correct options are
A

2

B

1

D

3

Any point on the line $x+y=1$ can be taken as $(t,1-t)$.

Equation of the chord, with this as mid-point is

$y(1-t)-2a(x+t)=(1-t{)}^2-4at$, it passes through (a, 2a).

So $t^2-2t+2a^2-2a+1=0$,

this should have 2 distinct real roots so discriminant > 0, we get $a^2-a<0$

$\Rightarrow 0<a<1$, so length of latus rectum < 4

$\Rightarrow$ latus rectum $\ne 4$.