Question

Find the coordinates of the point of intersection of the tangents to the graph of $y=x^2$ at the points at which it meets the line with equation $y=x+2$

Answer 1

Points of intersection of line and parabola:

has x-coordinates condition $x^2=x+2$

$\Rightarrow$ $x^2-x-2=0$

$\Rightarrow$ $x=\frac{1\pm \sqrt{1+8}}{2}=\frac{1\pm 3}{2}$

$=2$ or $-1$

$\Rightarrow$ $y=4$ or $1$

$\Rightarrow$ POI are $(2,4),(-1,1)$

Slo[e of tangents $\frac{dy}{dx}=\frac{d}{dx}\left(x^2\right)=2x$

At $(2,4)$ slope $=4$, line dquation $y-4=4(x-2)$

$\Rightarrow$ $y=4x-4$

At $(-1,1)$ slope $=-2$ line equation $y-1=-2(x+1)\Rightarrow$$y=-2x-1$

For POI, $4x-4=-2x-1$

$\Rightarrow$ $6x=3$

$\Rightarrow$ $x=\frac{1}{2}$

$\Rightarrow$ $y=4\left(\frac{1}{2}\right)=-2$

$\Rightarrow$ POI$=(\frac{1}{2},-2)$