Question

The normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the axes in $M$ and $N$ and the lines $MP$ and $NP$ are drawn at right angles to the axes. Prove that the locus of $P$ is the hyperbola $a^2{x}^2-b^2{y}^2={(a^2+b^2)}^2$

Answer 1

Normal to hyperbola at $(x_1,y_1)$ on hyperbola is $a^{2}x{x}_1+b^{2}y/y_1=a^2{e}^2$

$M(x_1{e}^2,0)$

$N(O,\frac{a^2{e}^2}{b^2}{y}_1)$

$P(x_1{e}^2,\frac{a^2{e}^2}{b^2}{y}_1)$

$x=x_1{e}_1^2,y=\frac{a^2{e}^2}{b^2}{y}_1$

$\Rightarrow x_1=\frac{x}{e_1^2},y_1=\frac{a^2{b}^2}{a^2{b}^2}$

$\Rightarrow x_1=\frac{a{a}^2}{a^2+b^2},y_1=\frac{a^2{b}^2}{a^2+b^2}$

$\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1\Rightarrow \frac{a^4{x}^2}{(a^2+b^2{)}^2{a}^2}-\frac{y^2{b}^4}{(a^2+b^2{)}^2{b}^2}=1$

$\Rightarrow a^2{x}^2-b^2{y}^2=(a^2+b^2{)}^2$