Question

if x=a cos theta , y=b sin theta then find value of b^2x^2+a^2y^2-a^2b^2

Answer 1

Answer:

Given

x = acosA , y = bsinA

We have

b^2 * x^2 + a^2 * y^2 - a^2 * b^2 ......................(i)  

put x & y values in equation (i)

b^2 a^2 cos^2 A + a^2 b^2 sin^2 A - a^2 * b^2

a^2 b^2 (cos^2 A +sin^2 A -1)

a^2 b^2 (1-1)               [sin^2 A + cos^2 A = 1]

a^2 b^2 * 0

= 0

thank you.................

Answer 2

Step-by-step explanation:

x=acosθ;y=bsinθ

cosθ= ax ;sinθ= y

cos 2 θ+sin 2 θ=1= a 2x 2 + b 2y 2

x 2 b 2 +a 2 y 2 =a 2 b 2 .