Question

A tangent to the hyperbola $x^2-2y^2=4$ meets x-axis at P and Y-axis at 'Q'. Line PR and QR are drawn such that OPRQ is a rectangle (where O is origin). The locus of R is:

• A. $\frac{4}{x^2}+\frac{2}{y^2}=1$
• B. $\frac{4}{x^2}-\frac{2}{y^2}=1$
• C. $\frac{2}{x^2}+\frac{4}{y^2}=1$
• D. $\frac{2}{x^2}-\frac{4}{y^2}=1$

Answer 1

The correct option is B $\frac{4}{x^2}-\frac{2}{y^2}=1$

Solution -

Equation of tangent to hyperbola $\frac{x^2}{4}-\frac{4^2}{2}=1$

is $\frac{xsec\theta }{2}-\frac{ytan\theta }{\sqrt{2}}=1$ at any parametric point $\theta$

P is $(\frac{2}{sec\theta },0)$

Q is $(0,\frac{-\sqrt{2}}{tan\theta })$

R will be x coordinate of r is taken and y - coordinate

Q is taken

$R(\frac{2}{sec\theta },\frac{-\sqrt{2}}{tan\theta })$ $h=\frac{2}{sec\theta }$ $K=\frac{-\sqrt{2}}{tan\theta }$

$\frac{4}{h^2}=\frac{2}{k^2}=1\to \frac{4}{x^2}-\frac{2}{y^2}=1$

B is correct.