Question

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

Solving the two equations, we getx²y = xy

Ž xy (x – 1) = 0 Ž x = 0, y = 0, x = 1. Since y = 0 does not satisfy the two equations. So, we neglect it. Putting

x = 0 in the either equation, we get x = 1. Now, putting

x = 1 in one of the two equations we obtain y = 1/2.

Thus, the two curves intersect at (0, 1) and (1, 1/2).

Now, x²y = 1 – y

Ž x² (dy/dx) + 2xy = – (dy/dx)

Ž dy/dx = –(2xy)/(x² + 1).

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

The equations of the required tangents are

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

Ž y = 1 and x + 2y – 2 = 0

These two tangents intersect at (0, 1)