$\mathrm{A}\equiv \left(-4,0\right),\mathrm{B}\equiv \left(4,0\right).$ M and N are the variable points of the y-axis such that M lies below N and $\mathrm{MN}=4.$ Lines AM and BN intersect at P. The locus of P is 

Let $\mathrm{M}\equiv \left(0,\mathrm{h}\right).$ Then 

or $\mathrm{N}\equiv \left(0,\mathrm{h}+4\right)$

The equation of AM is 

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

or $\frac{\mathrm{y}}{\mathrm{h}}+\frac{4+\mathrm{x}}{4}$

or $\mathrm{h}=\frac{4\mathrm{y}}{4+\mathrm{x}}$

The equation of BN is 

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

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

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

or $\mathrm{h}=\frac{4\mathrm{y}-16+4\mathrm{x}}{4-\mathrm{x}}$

or $\frac{4\left(\mathrm{y}-4+\mathrm{x}\right)}{4-\mathrm{x}}=\frac{4\mathrm{y}}{4+\mathrm{x}}$ ( Eliminating h )

or $2\mathrm{xy}-16+{\mathrm{x}}^2=0,$ which is the required locus