This is the question ?

x/acostheta+y/bsintheta=1
x/asintheta-y/bcostheta=1
Then prove that
x^2/a^2+y^2/b^2=2

2 Answers
Jun 1, 2018

Given,

rarr(x/a)costheta+(y/b)sintheta=1.....(1)

rarr(x/a)sintheta-(y/b)costheta=1......(2)

Squaring and adding equation (1) and (2),we get

rarr(x^2)/(a^2)cos^2thetacancel(+2((xy)/(ab))costheta*sintheta)+(y^2)/(b^2)sin^2theta+(x^2)/(a^2)sin^2thetacancel(-2((xy)/(ab))sintheta*costheta)+(y^2)/(b^2)cos^2theta=1^2+1^2

rarr(x^2)/(a^2)(cos^2theta+sin^2theta)+(y^2)/(b^2)(sin^2theta+cos^2theta)=2

therefore(x^2)/(a^2)+(y^2)/(b^2)=2

Jun 1, 2018

Please see below.

Explanation:

Here,

x/acostheta+y/bsintheta=1to(1)

x/asintheta-y/bcostheta=1to(2)

Squaring (1) and (2) and then adding we get

x^2/a^2cos^2theta+2(xy)/(ab)costhetasintheta+y^2/b^2sin^2 theta=1

ul(x^2/a^2sin^2theta- 2(xy)/(ab)costhetasintheta+y^2/b^2cos^2theta=1)

x^2/a^2(cos^2theta+sin^2theta)+0+y^2/b^2(sin^2theta+cos^2 theta)=1+1

i.e. x^2/a^2(1)+y^2/b^2(1)=2

Hence,

x^2/a^2+y^2/b^2=2