Question

Eliminating $\theta$ from the following equation $x=a(\sec \theta +\tan \theta ),y=b(\sec \theta -\tan \theta )$ then

• A. $xy=ab$
• B. $x^2+y^2=a^2+b^2$
• C. $x^2-y^2=a^2-b^2$
• D. none of these

Answer 1

The correct option is A $xy=ab$

Given,

$x=a(\sec \theta +\tan \theta ),y=b(\sec \theta -\tan \theta )$

or, $\sec \theta =\frac{x}{a}$......(1), $\sec \theta -\tan \theta =\frac{y}{b}$.....(2).

We've,

${\sec }^2\theta -{\tan }^2\theta =1$

or, $(\sec \theta -\tan \theta )(\sec \theta -\tan \theta )=1$

or, $\frac{x}{a}.\frac{y}{b}=1$ [ Using (1) and (2)]

or, $xy=ab$.