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SSC CHSL 6 Dec 2015 Shift 1 Paper

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Question : 116 of 200

Marks: +1, -0

If

x=a(\sin\theta+\cos\theta)

y==a(\sin\theta-\cos\theta)

, then the value of

\frac{x^2}{a^2}+\frac{y^2}{b^2}

is:

3
4
2
1

Solution:

Given :

x=a(\sin\theta+\cos\theta)

and

y==a(\sin\theta-\cos\theta)

Squaring both sides, we get :

⇒

x^2 = a^2(\sin\theta + \cos\theta)^2

⇒

x^2 = a^2(\sin^2\theta + \cos^2\theta

+ 2\sin\theta \cos\theta)

⇒

x^2 = a^2(1 + 2\sin\theta \cos\theta)

⇒

\frac{x^2}{a^2}=1+2\sin\theta\cos\theta

----(i)

⇒ Similarly,

\frac{y^2}{b^2}=1-2\sin\theta\cos\theta

----(ii)

Adding both equations (i) and (ii),

⇒

\frac{x^2}{a^2}+\frac{y^2}{b^2}=(1+2\sin\theta\cos\theta)

+(1-2\sin\theta\cos\theta)

=

1+1=2

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