Question

If $x=a\sec \theta \cos \varphi ,y=b\sec \theta \sin \varphi$ and $z=c\tan \theta$ then prove that $(\frac{x^2}{a^2}+\frac{y^2}{b^2})=(1+\frac{z^2}{c^2})$.

Answer 1

We have,

$x=a\sec \theta \cos \varphi$

$\frac{x}{a}=\sec \theta \cos \varphi$

On squaring both sides, we get

$\frac{x^2}{a^2}={\sec }^2\theta {\cos }^2\varphi$ ……. (1)

Now, $y=b\sec \theta \sin \varphi$

$\frac{y}{b}=\sec \theta \sin \varphi$

On squaring both sides, we get

$\frac{y^2}{b^2}={\sec }^2\theta {\sin }^2\varphi$ ……. (2)

Now, $z=c\tan \theta$

$\frac{z}{c}=\tan \theta$

On squaring both sides, we get

$\frac{z^2}{c^2}={\tan }^2\theta$ ……. (3)

On adding equation (1) and (2), we have,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}={\sec }^2\theta {\cos }^2\varphi +{\sec }^2\theta {\sin }^2\varphi$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}={\sec }^2\theta ({\cos }^2\varphi +{\sin }^2\varphi )$

We know that

${\cos }^2\varphi +{\sin }^2\varphi =1$

Thus,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}={\sec }^2\theta$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}={\tan }^2\theta +1$ ……. (4)

From equation (3) and (4), we get

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1+\frac{z^2}{c^2}$

Hence, proved.