Question

if x= acos theta and y =b sine theta then find the value of b2x2+a2y2-a2b2. (the 2 is a square).. que .9
plxxxx

Answer 1

Hey user !!

Here's the answer you are looking for

x = acosθ

y = bsinθ

So,

b²x² + a²y² - a²b²

= b² (acosθ)² + a² (bsinθ)² - a²b²

= b²a²cos²θ + a²b²sin²θ - a²b²

= a²b²cos²θ + a²b²sin²θ - a²b²

= a²b² ( cos²θ + sin²θ - 1)

= a²b² (1 - 1) [cos²θ + sin²θ = 1]

= a²b² (0)

= 0 (ans.)


★★ HOPE THAT HELPS ☺️ ★★

Answer 2

hello dude,



=> acosθ

y = bsinθ

So,

b²x² + a²y² - a²b²

= b² (acosθ)² + a² (bsinθ)² - a²b²

= b²a²cos²θ + a²b²sin²θ - a²b²

= a²b²cos²θ + a²b²sin²θ - a²b²

= a²b² ( cos²θ + sin²θ - 1)

= a²b² (1 - 1) [cos²θ + sin²θ = 1]

= a²b² (0)

= 0
hope its help you:-)