Question

Find the locus of mid-point of chord of
parabola $y^2=4x$ which touches the parabola $x^2=4y$.

• A. $x^3+2xy+4=0$
• B. $y^3+2xy+4=0$
• C. $x^3-2xy+4=0$
• D. $y^3-2xy+4=0$

Answer 1

The correct option is D

$y^3-2xy+4=0$

Let (h,k) be the point of chord of parabola $y^2=4x$

so,equation of chord is

$T=S_1$

$y{y}_1-2a(x+x_1)=y_1^2-4a{x}_1$

$y\left(k\right)-2a(x+h)=k^2-4h$

$ky-2x-k^2+4h-2h=0$

$ky-2x-k^2+2h=0$ - - - - - - - - (1)

This chord touches the parabola $x^2=4y$

It means given chord is a tangent to the parabola

$x^2=4y$

Point of contact of chord of contact

$ky-2x-k^2+2h=0$ & parabola $x^2=4y$ is

substituting $y=\frac{x^2}{4}$ in equation (1)

$k.\frac{x^2}{4}-2x-k^2+2h=0$

$\frac{k}{4}{x}^2-2x-k^2+2h$

since,it touches the parabola;there should be only one root

discriminant(D)=0

$b^2-4ac=0$

$4-4\times \left(\frac{k}{4}\right).(-k^2+2h)=0$

$1-\left(\frac{k}{4}\right)(-k^2+2h)=0$

$\frac{k}{4}(-k^2+2h)=4$

For required locus replacing $k=y\&h=x$

we get

$y(-y^2+2x)=4$

$-y^3+2xy=4$

$y^3-2xy+4=0$