Question : If $x=a\left ( \sin heta+\cos heta ight )$ and $y=b\left ( \sin heta-\cos heta ight )$, then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is:
Option 1: 4
Option 2: 3
Option 3: 1
Option 4: 2

Question

Question : If $x=a\left ( \sin heta+\cos heta ight )$ and $y=b\left ( \sin heta-\cos heta ight )$, then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is:

Option 1: 4

Option 2: 3

Option 3: 1

Option 4: 2

Team Careers360 · Accepted Answer

Correct Answer: 2

Solution : Given: $x=a\left ( \sin heta+\cos heta ight )$ and $y=b\left ( \sin heta-\cos heta ight )$.
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = \frac{a^{2}\left ( \sin^{2} heta+2\sin heta\cos heta+\cos^{2} heta ight )}{a^{2}} + \frac{b^{2}\left ( \sin^{2} heta-2\sin heta\cos heta+\cos^{2} heta ight )}{b^{2}}$
⇒ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = \left ( \sin^{2} heta+2\sin heta\cos heta+\cos^{2} heta ight ) + \left ( \sin^{2} heta-2\sin heta\cos heta+\cos^{2} heta ight )$
⇒ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 2\sin^{2} heta+2\cos^{2} heta$
⇒ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 2(\sin^{2} heta+\cos^{2} heta)$
Since $\sin^{2} heta+\cos^{2} heta=1$,
⇒ $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2$
Hence, the correct answer is 2.

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Question : If x=a(sin⁡θ+cos⁡θ) and y=b(sin⁡θ−cos⁡θ), then the value of x2a2+y2b2 is:Option 1: 4Option 2: 3Option 3: 1Option 4: 2

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Question : If $x = a (\sin θ + \cos θ)$ and $y = b (\sin θ - \cos θ)$ , then the value of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ is:
Option 1: 4
Option 2: 3
Option 3: 1
Option 4: 2