Question

If x=a cosA and y=b sinA.Find b^2x^2+a^2y^2-a^2b^2

Answer 1

b^2x^2+a^2y^2+a^2b^2.
put value x=a cosA , y=bsinA
b^2a^2 cos^2A + a^2b^2 sin^2A - a^2b^2
a^2b^2 ( cos^2A+ sin^2A - 1 )
: put 【sin^2A + cos^2A=1】
a^2b^2 ( 1 - 1 )
a^2b^2 (0)
0

Answer 2

Hi ,

x = a cosA

x /a = cosA ----( 1 )

y = b sinA

y / b = sinA ----( 2 )

( cosA + sinA )² = cos² A + sin² A + 2cosAsinA

( x/a + y/b )² = 1 + 2 cosAsinA

[since cos² A + sin² A = 1 ]

[ from ( 1 ) and ( 2 ) ]

(x/a)² + ( y/b)² + 2 xy/ab = 1 + 2xy/ab

x²/a² + y² / b² = 1

multiply each term with a² b² , we get

b² x²+ a² y² = a² b²

Therefore ,

b²x² + a² y² - a² b² = 0

I hope this helps you.

:)