Question

if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2  = 2`

Answer 1

`[x/a cos theta + y/b sin theta]^2 + [x/a sin theta - y/b cos theta] = (1)^2 + (1)^2`

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

`x^2/a^2  cos^2 theta + y^2/b^2 cos^2 theta  + y^2/b^2 sin^2  theta  = 2` 

`cos^ theta [x^2/a^2 + y^2/b^2] + sin^2 theta(x^2/a^2  + y^2/a^2) = 2`

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

Answer 2

It is given that:

`x/a cos θ + y/b sin θ = 1`     ....(A)

and `x/a sin θ - y/b cos θ = 1`    ....(B) 

On squaring equation (A), we get

`(x/a cos θ + y/b sin θ)^2 = (1)^2`

⇒ `x^2/a^2 cos^2 θ + y^2/b^2 sin^2 θ + 2 x/a . y/b sin θ. cos θ = 1`     ....(c)

On squaring equation (B), we get

= `(x/a sin θ - y/b cos θ )^2 = (1)^2`

⇒ `x^2/a^2 sin^2 θ + y^2/b^2 cos^2 θ + 2 x/a . y/b sin θ. cos θ = 1`  ....(D)

Adding (C) and (D), we get,

⇒ `x^2/a^2 cos^2 θ + y^2/b^2 sin^2 θ + 2 x/a . y/b sin θ. cos θ + x^2/a^2 sin^2 θ + y^2/b^2 cos^2 θ + 2 x/a . y/b sin θ. cos θ = 1 + 1`

⇒ `x^2/a^2 sin^2 θ  + y^2/b^2cos^2 θ-(4xy)/"ab" sin^2 θ + cos^2 θ = 2`

⇒ `x^2/a^2 xx 1 + y^2/b^2 xx 1 = 2`

⇒ `x^2/a^2 + y^2/b^2 = 2`

Hence proved.