Question

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

Answer 1

We know that, the equation of line is given as,

$y-y_1=m(x-x_1)$

Equation of line with slope $1$ and from point $A(-3,0)$

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

$y=x+3$

$x-y+3=0$

Above line cuts the curve $y=x^2+x+1$

On solving curve and line

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

$x^2=2$

$x=\pm \sqrt{2}$

$\Rightarrow y=\sqrt{2}+3$ and $y=3-\sqrt{2}$

$\Rightarrow P(\sqrt{2},3+\sqrt{2})$ and $Q(-\sqrt{2},3-\sqrt{2})$

We know that,

$distance=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$l\left(AP\right)=\sqrt{(-3-\sqrt{2}{)}^2+(0-3-\sqrt{2}{)}^2}$

$l\left(AP\right)=\sqrt{(-3-\sqrt{2}{)}^2+(-3-\sqrt{2}{)}^2}$

$l\left(AP\right)=\sqrt{2(-3-\sqrt{2}{)}^2}$

$l\left(AP\right)=(-3\sqrt{2}-2)$

$l\left(AQ\right)=\sqrt{(-3+\sqrt{2}{)}^2+(0-3+\sqrt{2}{)}^2}$

$l\left(AQ\right)=\sqrt{(-3-\sqrt{2}{)}^2+(-3+\sqrt{2}{)}^2}$

$l\left(AQ\right)=\sqrt{2(-3+\sqrt{2}{)}^2}$

$l\left(AQ\right)=(-3\sqrt{2}+2)$

$l\left(AP\right)l\left(AQ\right)=-(2+3\sqrt{2})(2-3\sqrt{2})$

$l\left(AP\right)l\left(AQ\right)=-(4-6)$

$l\left(AP\right)l\left(AQ\right)=2$