MCQ
If $\text{x}=\text{a}\cos\theta\text{ and y}=\text{b}\sin\theta,$ then $b^2x^2 + a^2y^2 =$
- ✓$a^2b^2$
- B$ab$
- C$a^4b^4$
- D$a^2 + b^{_2}$
✓
Answer
Correct option: A.
$a^2b^2$
$a^2b^2$
$\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta\ .....(\text{i})$
$\text{bx}=\text{ab}\cos\theta,\text{ay}=\text{ab}\sin\theta\ .....(\text{ii})$
Adding (i) and (ii) we get,
$=\text{b}^2\text{x}^2+\text{a}^2\text{b}^2\cos^2\theta+\text{a}^2\text{b}^2\sin^2\theta$
$=\text{a}^2\text{b}^2(\cos^2\theta+\sin^2\theta)$
$=\text{a}^2\text{b}^2\times1$
$=\text{a}^2\text{b}^2$
$\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta\ .....(\text{i})$
$\text{bx}=\text{ab}\cos\theta,\text{ay}=\text{ab}\sin\theta\ .....(\text{ii})$
Adding (i) and (ii) we get,
$=\text{b}^2\text{x}^2+\text{a}^2\text{b}^2\cos^2\theta+\text{a}^2\text{b}^2\sin^2\theta$
$=\text{a}^2\text{b}^2(\cos^2\theta+\sin^2\theta)$
$=\text{a}^2\text{b}^2\times1$
$=\text{a}^2\text{b}^2$
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