Question

If `x/(a cosθ) = y/(b sinθ)   "and"  (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that"  x^2/a^2 + y^2/b^2 = 1`

Answer 1

Let `x/(acosθ) = y/(bsinθ)` ..............(1)

and `(ax)/(cosθ) - (by)/(sinθ) = a^2 - b^2` ...........(2)

From (1),  `y/sinθ = (xb)/(a cosθ)`

Putting (2) , we get `(ax)/cosθ - b (xb)/(acosθ) = a^2 - b^2`

⇒ `(ax)/cosθ - (xb^2)/(a cosθ) = a^2 - b^2`

⇒ `(x(a^2 - b^2))/(acosθ) = a^2 - b^2`

⇒ x = a cosθ

By(1), `y = (xb sinθ)/(a cosθ) = (a cosθ  bsinθ)/(a cosθ) = b sinθ`

Thus , `x^2/a^2 + y^2/b^2 = (a^2 cos^2θ)/a^2 + (b^2 sin^2θ)/b^2 = sin^2θ + cos^2θ = 1`