Question

The line x+y=2 is tangent to the curve $x^2=3-3y$ at the point

• A. (1, 2/3)
• B. (–1, 1)
• C. $(\sqrt{3},0)$
• D. (3, –3)

Answer 1

The correct option is A (1, 2/3)

Given curve $x^2=3-2y$
Differentiate w.r.t. x, $2x=0-2\frac{dy}{dx}\Rightarrow fracdydx=-x$
Slope of the tangent of the curve =-x
From the given line, slope =-1, $\therefore x=1$ and from equation (i), y=2/3
$\therefore$ Co-ordinate of the point is (1,2/3).