Question

The point of intersection of the tangents drawn to the curve x2y = 1–y at the points where it is meet by the curve

xy = 1–y, is given by

• A. (0, –1)
• B. (1, 1)
• C. (0, 1)
• D. None of these

Answer 1

Answer: (C)

(0, 1)

Answer 2

x²y = xy ⇒ xy (x – 1) = 0 ⇒ x = 0, y = 0, x = 1

Q y ≠ 0, so point of intersection of two curves are (0, 1) and (1, 1/2)

x²y = 1 – y ⇒ x² $\frac{dy}{dx}$ + 2xy = – $\frac{dy}{dx}$

⇒ $\frac{dy}{dx}$= – $\frac{2xy}{x^{2} + 1}$

$\left( \frac{dy}{dx} \right)_{(0,1)}$ = 0 and $\left( \frac{dy}{dx} \right)_{(1,1/2)}$ = – $\frac{1}{2}$

equation of tangent

(y – 1) = 0 (x – 0) and y – 1/2 = – 1/2 (x – 1)

y = 1 and x + 2y – 2 = 0

this intersect at (0, 1)