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`

Answer 1

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`

𝑆𝑞𝑢𝑎𝑟𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒, 𝑤𝑒 𝑔𝑒𝑡:

`(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     [∵ sin^2 theta + cos^2 theta =1]`

∴`x^2/a^2 + y^2/b^2 = 2`