Question
If 2ycosθ = x sinθ and 2xsecθ – ycosecθ = 3, then
x 2 + 4y 2 = ?Solution
We have, 2y cosθ = xsinθ and 2xsecθ – y cosecθ =3
Put θ = 45º , then, 2y/√2 = x/√2
⇒2y = x ⇒ 2y – x = 0 …… (1)
And, 2x√2 – y√2 = 3
⇒2x – y = 3/√2………… (2)
Now, (ii) + 2 (i) we get 2x – y = 3/√2
–2x + 4y = 03y = 3/√2
3y = 3/√2 ⇒ y = 1/√2 So, x = 2y = 2 × 1/√2 = 2
=2 × 1/√2 = 2
Now, x 2 + 4y 2 ⇒ (√2) 2 + 4 × (1√2) 2 ⇒ 2 + 2 = 4
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