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Question
If `(x/a sin a - y/b cos theta) = 1 and (x/a cos theta + y/b sin theta ) =1, " prove that "(x^2/a^2 + y^2/b^2 ) =2`
Solution
We have `(x/a sin theta - y/a cos theta ) =1`
Squaring both side, we have:
`(x/a sin theta - y/b cos theta )^2 = (1)^2`
⇒ `(x^2/a^2 sin^2 theta + y^2/b^2 cos^2 theta - 2 x/a xx y/b sin theta cos theta ) = 1 .....(i)`
Again , `(x/a cos theta + y/b sin theta ) =1`
๐๐๐ข๐๐๐๐๐ ๐๐๐กโ ๐ ๐๐๐, ๐ค๐ ๐๐๐ก:
`(x/a cos theta + y/b sin theta )^2 = (1)^2`
`⇒ (x^2/a^2 cos^2 theta + y^2 /b^2 sin ^2 theta + 2 x/a xx y/b sin theta cos theta ) = ....(ii)`
Now, adding (i) and (ii), we get:
`(x^2/a^2 sin^2 theta + y^2 /b^2 cos^2 theta -2 x/a xx y/b sin theta cos theta ) + (x^2/a^2 cos^2 theta + y^2 / b^2 sin^2 theta + 2 x/a xx y/b sin theta cos theta)`
⇒`x^2/a^2 sin^2 theta + y^2/b^2 cos^2 theta + x^2 /a^2 cos^2 theta + y^2/b^2 sin^2 theta =2`
⇒`(x^2/a^2 sin^2 theta + x^2/a^2 cos^2 theta)+(y^2/b^2 cos^2 theta + y^2/b^2 sin ^2 theta ) =2`
⇒`x^2/a^2 (sin^2 theta + cos^2 theta ) + y^2/b^2 (cos^2 theta + sin^2 theta ) =2`
⇒`x^2/a^2 + y^2 /b^2 =2 [โต sin^2 theta + cos^2 theta =1]`
∴`x^2/a^2 + y^2/b^2 = 2`
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