Question

Answer Answer !!!!!​

Answer 1

Answer:

Step-by-step explanation:

Work 1:

Given, x = a cos theta and y = b sin theta. Substituting for x and y ,

b²x² + a²y² - a²b² = b²a² cos² theta + a²b² sin² theta - a²b²

= a²b² (cos² theta + sin² theta) - a²b²

= a²b² . 1 - a²b² (Using the trigonometric identity cos² theta + sin² theta = 1)

= a²b² - a²b² = 0 (Proved)

Work 2:

x = a cos theta , y = b sin theta . Substituting for x,

b²x² + a²y² - a²b² = b²a² cos² theta - a²b² + a²y²

= b²a²(cos² theta-1) + a²y²

= b²a²(-sin² theta) + a²y² = -a²(b²sin² theta) + a²y² Substitute y for b sin theta.

=-a²y² + a²y²

= 0 (Proved)

Answer 2

Given that :

$x = a \cos \alpha \: and \: y = b \sin\alpha$

Then,

${b}^{2} {x}^{2} + {a}^{2} {y}^{2} \\ \\ = > {b}^{2} {a}^{2} { \cos }^{2} \alpha + {a}^{2} {b}^{2} { \sin}^{2} \alpha \\ \\ = > {a}^{2} {b}^{2} ( { \sin }^{2} \alpha + { \cos }^{2} \alpha ) \\ \\ = > {a}^{2} {b}^{2} \times 1 \\ \\ = > {a}^{2} {b}^{2}$

So, Option (b) is correct ✔️✔️

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