Question

The normal to the hyperbola $4x^2-9y^2=36$ meets the axes in $M$ and $N$ and the lines $MP$, $NP$ are drawn right angles at the axes. The locus of $P$ is the hyperbola

• A. $9x^2-4y^2=169$
• B. $4x^2-9y^2=169$
• C. $3x^2-4y^2=169$
• D. $Noneofthese$

Answer 1

Answer: (D)

$Noneofthese$

$\frac{x^2}{9}-\frac{y^2}{4}=1$.Let $P(x_1,y_1)$ be the point on hyperbola.

Eqn of the normal is$\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2{b}^{2}M=\left(\frac{a^2-b^2}{a^2}\right)x_1=xN=\left(\frac{a^2-b^2}{a^2}\right)y_1=yP=(x,y)$$x_1=\frac{a^2\left(x\right)}{a^2-b^2}$ $(x_1,y_1)$ lies at hyperbola.$y_1=\frac{b^2\left(y\right)}{a^2-b^2}$ $(x_1,y_1)$ lies at hyperbola.Now, $a^2=9,b^2=4$Therefore, $x_1=\frac{9x}{5},y_1=\frac{4y}{5}$$\frac{x_1^2}{9}-\frac{y_1^2}{4}=1{\left(\frac{9x}{5}\right)}^2\frac{x_1^2}{9}-{\left(\frac{4y}{5}\right)}^2\frac{y_1^2}{4}=1⟹9x^2-4y^2=25$