Question

If `sqrt(1 - x^2) + sqrt(1 - y^2) = "a"(x - y)`, prove that `"dy"/"dx" = sqrt((1 - y^2)/(1 - x^2)` 

Answer 1

Given that: `sqrt(1 - x^2) + sqrt(1 - y^2) = "a"(x - y)`

Put x = sin θ and y = sin Φ.

∴ θ = sin^–1x and Φ = sin^–1y

`sqrt(1 - sin^2theta) + sqrt(1 - sin^2phi)` = a(sin θ – sin Φ)

⇒ `sqrt(cos^2theta) + sqrt(cos^2phi)` = a(sin θ – sin Φ)

⇒ cos θ + cos Φ = a(sin θ – sin Φ)

⇒ `(cos theta + cos phi)/(sin theta - sin phi)` = a

⇒ `(2 cos  (theta + phi)/2 * cos  (theta - phi)/2)/(2cos  (theta + phi)/2 * sin  (theta - phi)/2)` = a        ......`[(because cos "A" + cos "B" = 2cos  ("A" + "B")/2 * cos  ("A" - "B")/2),(sin"A" - sin"B" = 2cos  ("A" + "B")/2 * sin  ("A" - "B")/2)]`

⇒ `(cos((theta - phi)/2))/(sin((theta - phi)/2))` = a

⇒ `cot((theta - phi)/2)` = a

⇒ `(theta - phi)/2 = cot^-1"a"`

⇒ θ – Φ = 2cot^–1a

⇒ sin^–1x – sin^–1y = 2 cot^–1a

Differentiating both sides w.r.t. x

`"d"/"dx" (sin^-1x) - "d"/"dx"(sin^-1x) = 2*"d"/"dx" cot^-1"a"`

⇒ `1/sqrt(1 - x^2) - 1/sqrt(1 - y^2) * "dy"/"dx"` = 0

⇒ `1/sqrt(1 - y^2) * "dy"/"dx" = 1/sqrt(1 - x^2)`

∴ `"dy"/"dx" = sqrt(1 - y^2)/sqrt(1 - x^2)`.