Question

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

• A. $(4,4)$
• B. $(-1,2)$
• C. $(\frac{9}{4},\frac{3}{8})$
• D. $(0,0)$

Answer 1

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

$y^2=x(2-x{)}^2$
$\Rightarrow 2y\left(\frac{dy}{dx}\right)=(2-x{)}^2-2x(2-x)$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,1)}=-\frac{1}{2}$
Hence, the equation of the tangent is
$y-1=-\frac{1}{2}(x-1)$
$\Rightarrow 2y-2=-x+1\Rightarrow x+2y=3$
On solving, the given curve and the tangent, we get
$y^2=(3-2y)(2-3+2y{)}^2$
$\Rightarrow y^2=(3-2y)(2y-1{)}^2$
$\Rightarrow 8y^3-19y^2+14y-3=0$
$\Rightarrow (y-1)(8y^2-11y+3)=0$
$(y-1{)}^2(8y-3)=0$
$\Rightarrow y=1,\frac{3}{8}$
when $y=\frac{3}{8},x=3-2y=3-\frac{3}{4}=\frac{9}{4}$
Hence, the point is $(\frac{9}{4},\frac{3}{8})$