If x=a ,y=b and z=c , then x^2 a^2 + y^2 b^2 = - Vidyadip

MCQ

If $\text{x}=\text{a}\sec\theta\cos\phi,\text{y}=\text{b}\sec\theta\sin\phi$ and $z=\text{c}\tan\theta,$ then $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=$

A
$\frac{\text{z}^2}{\text{c}^2}$
B
$1-\frac{\text{z}^2}{\text{c}^2}$
C
$\frac{\text{z}^2}{\text{c}^2}-1$
✓
$1+\frac{\text{z}^2}{\text{c}^2}$

✓

Answer

Correct option: D.

$1+\frac{\text{z}^2}{\text{c}^2}$

$\text{x}=\text{a}\sec\theta\cos\phi$
$\text{y}=\text{b}\sec\theta\sin\phi$
$\text{z}=\text{c}\tan\theta$
$\frac{\text{x}}{\text{a}}=\sec\theta\cos\phi\ .....(\text{i})$
$\frac{\text{y}}{\text{b}}=\sec\theta\sin\phi\ .....(\text{ii})$
$\frac{\text{z}}{\text{c}}=\tan\theta\ .....(\text{iii})$
Squaring and adding $(i)$ and $(ii)$
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=\sec^2\theta\cos^2\phi+\sec^2\theta\sin^2\phi$
$=\sec^2\theta(\cos^2\phi+\sin^2\phi)$
$=\sec^2\theta\times1=\sec^2\theta$
Squaring $(iii)$ and subtracting from $(iv)$
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=\sec^2\theta-\tan^2\theta=1$
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1+\frac{\text{z}^2}{\text{c}^2}$

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