Question

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

Answer 1

Consider the given expression.

$y^2=x{(2-x)}^2.........................\left(1\right)$

Differentiating this equation with respect to x, and we get,

$\frac{d}{dx}{y}^2=\frac{d}{dx}\left(x{(2-x)}^2\right)$

$\frac{d}{dx}{y}^2=\frac{d}{dx}\left(x(4+x^2-4x)\right)$

$\frac{d}{dx}{y}^2=\frac{d}{dx}(4x+x^3-4x^2)$

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

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

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

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

Now put, $(x,y)=(1,1)$and we get,

Slope $m=$ $\frac{dy}{dx}=\frac{4+3{\left(1\right)}^2-8\left(1\right)}{2\left(1\right)}$

$m=$ $\frac{dy}{dx}=\frac{-1}{2}$

We know that equation of tangent is,

$y-y_1=\frac{dy}{dx}(x-x_1)$

If given point $(1,1)$ lies on this tangent, so

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

$2y-2=-x+1$

$x+2y-2-1=0$

$x+2y-3=\mathrm{0...................................}\left(2\right)$

$2y=3-x$

$y=\frac{3-x}{2}............................\left(3\right)$

Put the value of $y$ in equation (1),

$y^2=x{(2-x)}^2$

${\left(\frac{3-x}{2}\right)}^2=x{(2-x)}^2$

Taking square root on both sides, we get,

$\left(\frac{3-x}{2}\right)=\sqrt{x}(2-x)$

$3-x=(4-2x)\sqrt{x}$

Let us take $x=t^2....................\left(4\right)$

Then,

$3-t^2=(4-2t^2)\sqrt{t^2}$

$3-t^2=(4-2t^2)t$

$3-t^2=4t-2t^3$

$2t^3-t^2-4t+3=0$

Solve this by remainder theorem,

Put $t=1$ and solve

$2{\left(1\right)}^3-{\left(1\right)}^2-4\left(1\right)+3=0$

$2-1-4+3=0$

$1-1=0$

$0=0$

Then divide equation (4) by$(t-1)$

$2t^3-t^2-4t+3=0$this solution is $(t-1)(2t^2+t-3)=0$

Then solve$(t-1)(2t^2+t-3)=0$ factorise $(2t^2+t-3)=0$

$t=1,1,\frac{-3}{2}$

Put the value of t in equation (4),

$x={\left(\frac{-3}{2}\right)}^2=\frac{9}{4}$ for any point P

$x=\frac{9}{4}$ put value of x in equation $\left(2\right)$ and we get,

$\frac{9}{4}+2y-3=0$

$2y=3-\frac{9}{4}$

$2y=\frac{12-9}{4}$

$2y=\frac{3}{4}$

$y=\frac{3}{8}$ for any point P

Hence, $(x,y)=(\frac{9}{4},\frac{3}{8})$ for any point P

It is required solution.