Question

Find the locus of the middle points of chords of the parabola $y^2=4ax$ which are of constant length $2l$.

• A. $(4ax+y^2)(y^2+4a^2)=4a^3{l}^2$
• B. $(4ax-y^2)(y^2-4a^2)=4a^3{l}^2$.
• C. $(4ax-y^2)(y^2+4a^2)=4a^3{l}^2$.
• D. $(4ax+y^2)(y^2-4a^2)=-4a^3{l}^2$.

Answer 1

The correct option is C $(4ax-y^2)(y^2+4a^2)=4a^3{l}^2$.

$=>y=4ax.............\left(1\right)$

Let the midpoints of chord be $(h,k)$, now equation equation of chord whose midpoint is given as,

$=>ky-2a(x+h)=k^2-4ah$

$=>ky-2ax=k^2-2ah...................\left(2\right)$ $(m=\frac{2}{a}k,c=\frac{k^2-2ah}{k})$

Length of intercept at line $y=mx+c$ at parabola $y^2=4ax$ is

$L=\frac{4}{m^2}\sqrt{a(1+m^2(a-mc)}$

For equation $\left(2\right)$ given,

$2l=\frac{4}{({\frac{2a}{k})}^2}\sqrt{a(1+{\left(\frac{2a}{k}\right)}^2(a-\left(\frac{2a}{k}\right)(\frac{k^2-2ah}{k})}$

$=>2a^{2}l=k^2\sqrt{a\left(\frac{k^2+4a^2}{k^2}\right)\left(\frac{a{k}^2-2a{k}^2+4a^{2}h}{k^2}\right)}$

$=>4a^4{l}^2=a(k^2+4a^2)(-a{k}^2+4a^{2}h)$

$=>4a^3{l}^2=(k^2+4a^2)(4ah-k^2)$

For locus replacing $k\to y$ and $h\to x$,

$=>4a^3{l}^2=(y^2+4a^2)(4ax-y^2).$