Question

A line having slope $1$ is drawn from a point $A(-3,0)$ cuts a curve $y=x^2+x+1$ at $P$ and $Q$. Find $l\left(AP\right).l\left(AQ\right)$

Answer 1

Equation of line is

$y-0=1(x+3)$

$\Rightarrow y=x+3...\left(i\right)$

$y=x^2+x+1...\left(ii\right)$

$x+3=x^2+x+1$

$\Rightarrow x^2=2$

$\therefore x=\pm \sqrt{2}$

$A(-3,0),p(\sqrt{2},3+\sqrt{2}),Q(-\sqrt{2},3-\sqrt{2})$

$AP.AQ=\sqrt{(\sqrt{2}+3{)}^2+(3+\sqrt{2}{)}^2}.\sqrt{(3-\sqrt{2}{)}^2+(3-\sqrt{1}{)}^2}$

$=\sqrt{2}(3+\sqrt{2}).\sqrt{2}(3-\sqrt{2})$

$=2(9-2)$

$=2\times 7$

$=14$

$\therefore AP.AQ=14$