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
Abhishek K.
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#

maganbhai P.
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#

Related questions
Impact of this question
1292 views around the world
You can reuse this answer
Creative Commons License