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 at right angles to 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. None of these.

Answer 1

Answer: (D)

None of these.

$\frac{x^2}{9}-\frac{y^2}{4}=1$

Let $P(x_1,y_1)$ be the point on hyperbola

Equation of normal

$\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2-b^2$
$M=\left(\frac{a^2-b^2}{a^2}\right)x_1=x$
$N=\left(\frac{a^2-b^2}{a^2}\right)y_1=y$
$P=(x,y)$
$x_1=\frac{a^2\left(x\right)}{a^2-b^2}(x_1,y_1)$
$y_1=\frac{b^2\left(y\right)}{a^2-b^2}{a}^2=9,b^2=4$
$x_1=\frac{9x}{5},y_1=\frac{4y}{5}$
$\frac{x_1^2}{9}-\frac{y_1^2}{4}=1{\left(\frac{9x}{5}\right)}^2\times \frac{1}{9}-{\left(\frac{4y}{5}\right)}^2\times \frac{1}{4}=1$
$9x^2-4y^2=25$