Question

x=3sin theta,y=4cos theta √ 16x²+9y²​

Answer 1

ANSWER:-

Given:

⚫x= 3sin∅

⚫y= 4cos∅

To find:

$= > \sqrt{16 {x}^{2} + {9y}^{2} }$

From given hypothesis, we get;

x² = (3sin∅)² = 9sin²∅

y² = (4cos∅)² = 16 cos²∅

Therefore,

$= > \sqrt{ {16 x}^{2} + {9y}^{2} } \\ \\ = > \sqrt{16( 9 {sin {}^{2} \theta)} + 9(16 {cos}^{2} \theta) } \\ \\ = > \sqrt{144 {sin}^{2} \theta + 144 {cos}^{2} \theta } \\ \\ = > \sqrt{144( {sin}^{2} \theta + {cos }^{2} \theta } ) \\ \\ = > \sqrt{144} \: \: \: \: \: \: \: [ { sin }^{2} \theta + {cos}^{2} \theta = 1] \\ \\ = > 12$

Hope it helps ☺️