Question
If 2ycosĪø = x sinĪø and 2xsecĪø ā ycosecĪø = 3, then
x 2 + 4y 2 = ?Solution
We have, 2y cosĪø = xsinĪø and 2xsecĪø ā y cosecĪø =3
Put Īø = 45Āŗ , then, 2y/ā2 = x/ā2
ā2y = x ā 2y ā x = 0 ā¦ā¦ (1)
And, 2xā2 ā yā2 = 3
ā2x ā y = 3/ā2ā¦ā¦ā¦ā¦ (2)
Now, (ii) + 2 (i) we get 2x ā y = 3/ā2
ā2x + 4y = 03y = 3/ā2
3y = 3/ā2 ā y = 1/ā2 So, x = 2y = 2 Ć 1/ā2 = 2
=2 Ć 1/ā2 = 2
Now, x 2 + 4y 2 ā (ā2) 2 + 4 Ć (1ā2) 2 ā 2 + 2 = 4
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