Question

$A(-4,0),B(4,0)$ are two points. $M$ and $N$ are the variable points of $y-$axis such that $M$ lies below $N$ and $MN=4$. Line joining $AM$ and $BN$ intersect at $P$. Locus of $P$ is

• A. $2xy-16-x^2=0$
• B. $2xy+16-x^2=0$
• C. $2xy+16+x^2=0$
• D. $2xy-16+x^2=0$

Answer 1

The correct option is D $2xy-16+x^2=0$

Let $M=(0,h)$

Therefore $N$ will be $N=(0,h+4)$

Therefore, equation of $AM$ by using slope intercept form will be

$\frac{x}{-4}+\frac{y}{h}=1$

By rearranging the terms we get

Therefore $\frac{y}{h}=\frac{x+4}{4}$

$h=\frac{4y}{x+4}$ ...$\left(i\right)$

Similarly equation of $BN$ is

$\frac{x}{4}+\frac{y}{h+4}=1$

$\frac{4-x}{4}=\frac{y}{h+4}$

$h+4=\frac{4y}{4-x}$

$h=\frac{4y}{4-x}-4$ ...$\left(ii\right)$

Equating $\left(i\right)$ and $\left(ii\right)$ we get

$\frac{4y}{4-x}-4=\frac{4y}{4+x}$

$\frac{4y}{4-x}=\frac{4y}{4+x}+4$

now multiply the above equation with (4 - x).(4 + x)

we get $4y\times (4+x)=4y\times (4-x)+4\times (4+x)\times (4-x)$

so we get

$x^2+2xy-16=0$