Question

If $x=p\sec \theta$ and $y=q\tan \theta$, then

• A. $x^2-y^2=p^2{q}^2$
• B. $x^2{q}^2-y^2{p}^2=pq$
• C. $x^2{q}^2-y^2{p}^2=\frac{1}{p^2{q}^2}$
• D. $x^2{q}^2-y^2{p}^2=p^2{q}^2$

Answer 1

Answer: (D)

$x^2{q}^2-y^2{p}^2=p^2{q}^2$

$\frac{x}{p}=sec\theta$
$\frac{y}{q}=tan\theta$
And
$se{c}^2\theta -ta{n}^2\theta =1$
$\Rightarrow \frac{x^2}{p^2}-\frac{y^2}{q^2}=1$
$\therefore q^2{x}^2-p^2{y}^2=p^2{q}^2$