If x = a (sin 0 + cos 0), y = b (sin 0 - cos 0), then \(\frac{x^2}{a^2

1.	If x = a (sin 0 + cos 0), y = b (sin 0 - cos 0), then \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) =1. 12. 03. 24. 4
Answer» Correct Answer - Option 3 : 2 Given: x = a (sin 0 + cos 0) y = b (sin 0 - cos 0) Formula used: sin 0 = 0 cos 0 = 1 Calculation: x = a(0 + 1) = a y = b(0 - 1) = - b \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = \(\dfrac{a^2}{a^2}+\dfrac{(- b)^2}{b^2}\) ⇒ \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = 1 + 1 ∴ \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = 2

If x = a (sin 0 + cos 0), y = b (sin 0 - cos 0), then \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) =1. 12. 03. 24. 4

Answer» Correct Answer - Option 3 : 2

Given:

x = a (sin 0 + cos 0)

y = b (sin 0 - cos 0)

Formula used:

sin 0 = 0

cos 0 = 1

Calculation:

x = a(0 + 1) = a

y = b(0 - 1) = - b

\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = \(\dfrac{a^2}{a^2}+\dfrac{(- b)^2}{b^2}\)

⇒ \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = 1 + 1

∴ \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = 2