Question

Let a tangent to the curve $y^2=24x$ meet the curve $xy=2$ at the points $A$ and $B$. Then the mid points of such line segments $AB$ lie on a parabola with the

• A. length of latus rectum $\frac{3}{2}$
• B. length of latus rectum 2
• C. directrix $4x=3$
• D. directrix $4x=-3$

Answer 1

Answer: (C)

$y^2=24x$
$a=6$
$xy=2$
$AB\equiv ty=x+6t^2$.......(1)
$AB\equiv T=S_1$
$kx+hy=2hk$.........(2)
From (1) and (2)
$\frac{k}{1}=\frac{h}{-t}=\frac{2hk}{-6t^2}$
$\Rightarrow$ then locus is $y^2=-3x$
Therefore directrix is $4x=3$