Question

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

• A. $\left(-4,4\right)$
• B. $\left(1,-2\right)$
• C. $\left(\frac{9}{4},\frac{3}{8}\right)$
• D. None of these.

Answer 1

The correct option is C

$\left(\frac{9}{4},\frac{3}{8}\right)$

Explanation for the correct option.

Step 1: Find the slope of the tangent.

The given equation is

$\begin{array}{rcl}{y}^2& =& x{(2-x)}^2\\ & \Rightarrow & y^2=x\left(4+x^2-4x\right)\\ & \Rightarrow & y^2=4x+x^3-4x^2\end{array}$

The slope $m$ of the tangent will be $={\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}$.

So, by differentiating the given equation with respect to $x$, we get

$\begin{array}{rcl}2y\frac{dy}{dx}& =& 4+3x^2-8x\\ \Rightarrow \frac{dy}{dx}& =& \frac{4+3x^2-8x}{2y}\\ & \Rightarrow & {\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=\frac{4+3{\left(1\right)}^2-8\left(1\right)}{2\left(1\right)}\\ & =& -\frac{1}{2}\end{array}$

Step 2: Find the equation of the tangent.

Equation of tangent will be $y-y_1=m\left(x-x_1\right)$

The point $P$ lies on the tangent, so the equation of the tangent will be

$\begin{array}{rcl}y-1& =& -\frac{1}{2}\left(x-1\right)\\ & \Rightarrow & y+\frac{1}{2}x=\frac{1}{2}+1\\ \Rightarrow 2y+x& =& 3\\ \Rightarrow x& =& 3-2y\end{array}$

Step 3: Find $Q$.

Substituting the value of $x$, in the equation of curve, we get

$\begin{array}{rcl}{y}^2& =& \left(3-2y\right){(2-3+2y)}^2\\ \Rightarrow \left(3-2y\right){(-1+2y)}^2-y^2& =& 0\\ & \Rightarrow & \left(3-2y\right)(1+4y^2-4y)-y^2=0\\ \Rightarrow 8y^3-19y^2+14y-3& =& 0\\ \Rightarrow \left(y-1\right)\left(8y^2-11y+3\right)& =& 0\\ \Rightarrow {\left(y-1\right)}^2\left(8y-3\right)& =& 0\\ \Rightarrow y& =& 1\mathrm{or}\frac{3}{8}\end{array}$

Now, if $y=1,x=1$ and if $y=\frac{3}{8},x=\frac{9}{4}$

Hence, option C is correct.