If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =

Question:

If $x=a \sec \theta$ and $y=b \tan \theta$, then $b^{2} x^{2}-a^{2} y^{2}=$

(a) $a b$

(b) $a^{2}-b^{2}$

(c) $a^{2}+b^{2}$

(d) $a^{2} b^{2}$

Solution:

Given:

$x=a \sec \theta, y=b \tan \theta$

So,

$b^{2} x^{2}-a^{2} y^{2}$

$=b^{2}(a \sec \theta)^{2}-a^{2}(b \tan \theta)^{2}$

$=b^{2} a^{2} \sec ^{2} \theta-a^{2} b^{2} \tan ^{2} \theta$

$=b^{2} a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)$

We know that,

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

Therefore,

$b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2}$

Hence, the correct option is (d).