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. $(1,2)$

Answer 1

The correct option is C $(\frac{9}{4},\frac{3}{8})$

We have
$\Rightarrow y^2=x^3-4x^2+4x$ .....(i)

$\Rightarrow 2y\frac{dy}{dx}=3x^2-8x+4$

$\Rightarrow \frac{dy}{dx}=\frac{3x^2-8x+4}{2y}$

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{(1,1)}=\frac{3-8+4}{2}=-\frac{1}{2}$

The equation of the tangent at $(1,1)$ is

$-1=-\frac{1}{2}(x-1)$

$\Rightarrow x+2y-3=0$ .....(ii)

On solving Eqs. (i) and (ii) we get

$x=9/4$ and $y=3/8$

Hence,the coordinates of $P$ are $(\frac{9}{4},\frac{3}{8})$.