SSC CHSL 6 Dec 2015 Shift 1 Paper

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Question : 116 of 200

Marks: +1, -0

If

x=a(sinθ+cosθ)

and

y==a(sinθ-cosθ)

, then the value of

x^2/a^2+y^2/b^2

is:

3
4
2
1

Solution:

Given :

x=a(sinθ+cosθ)

and

y==a(sinθ-cosθ)

Squaring both sides, we get :
⇒

x^ 2 = a^ 2(sin θ + cos θ)^ 2

⇒

x^ 2 = a^ 2(sin^ 2 θ + cos ^2 θ

+ 2sin θ.cos θ)

⇒

x ^2 = a^ 2(1 + 2sin θ.cos θ)

⇒

x^2/a^2= 1+2sinθ.cosθ

----(i)
⇒ Similarly,

y^2/b^2= 1-2sinθ.cosθ

----(ii)
Adding both equations (i) and (ii),
⇒

x^2/a^2+y^2/b^2=(1+2sinθ.cosθ)

+(1-2sinθ.cosθ)

=

1 + 1 = 2

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