The locus of the midpointegrals of all chords of the parabol... | Mathematics | SolveeIt | Solveeit

Today, August 18

Question

The locus of the midpoints of all chords of the parabola $y^2 = 4ax$ through its vertex is another parabola with directrix

A

$x = - a$

B

$x = a$

C

$x = 0$

D

$x=-\frac{a}{2}$

Answer

$x=-\frac{a}{2}$

Explanation

Solution

$2h=at^{2}, 2k=2at \Rightarrow t-k/a, \Rightarrow 2h =a \frac{k^{2}}{a^{2}}, y^{2}=2ax$ , Equation of its directr $x=-a/2$