Question

A tangent to the parabola $x^2+4ay=0$ cuts the parabola $x^2=4by$ at A and B the locus of the mid point of AB is

• A. $\left(a+2b\right)x^2=4b^{2}y$
• B. $\left(b+2a\right)x^2=4b^{2}y$
• C. $\left(a+2b\right)y^2=4b^{2}x$
• D. $\left(b+2x\right)x^2=4a^{2}y$

Answer 1

The correct option is A $\left(a+2b\right)x^2=4b^{2}y$

Any tangent to the parabola $x^2=-4ay$ is

$x=my-a/m$

$mx-m{y}^2-a$

$mx-m^{2}y+a=\mathrm{0...}\left(1\right)$

Let $(x_1,y_1)$ be the mid point of A,B

$e{q}^n$ of chord AB

$S_1=T$

$x_1^2-4b{y}_1=x_{1}x-2b(y+y_1)$

$x_{1}x-2by-2b{y}_1+4b{y}_1-x_1^2=0$

$x_{1}x-2by+2b{y}_1-x_1^2=\mathrm{0...}\left(2\right)$

eq (1) and (2) represent the same line

$\frac{m}{x_1}=\frac{m^2}{2b}=\frac{a}{2b{y}_1-x_1^2}$

$m=\frac{2b}{x_1}$

also $\frac{m}{x_1}=\frac{a}{2b{y}_1-x_1^2}$

$4b^2{y}_1-2b{x}_1^2=a{x}_1^2$

$4b^2{y}_1=x_1^2(a+2b)$

so the locus in $4b^{2}y=x^2(a+2b)$