Question

If the tangent at (1, 1) on \(y^2 = x(2 - x)^2\) meets the curve again at P, then P is

• A. $(4,4)$
• B. $(–1,2)$
• C. $\frac{9}{4},\frac{3}{8})$
• D. $Noneofthese$

Answer 1

The correct option is B $(–1,2)$

$2^x+2^y=2^x+y\therefore 2^x.lo{g}_{e}2+2^y.lo{g}_e\frac{dy}{dx}=2^{x+y}.lo{g}_{e}2(1+\frac{dy}{dx})$
or $2^x+2^y\frac{dy}{dx}=2^{x+y}(1+\frac{dy}{dx})\therefore \frac{dy}{dx}=\frac{2^{x+y}-2^x}{2^y-2^{x+y}}\frac{dy}{dx}{|}_{x=y=1}=-1$