A line passing through the point A(-2,0), touches the parabola P:y^2=x-2 at the point B in…

Question

A line passing through the point $A(-2,0)$, touches the parabola $P:{y}^{2}=x-2$ at the point $B$ in the first quadrant. The area, of the region bounded by the line $AB$, parabola $P$ and the $x$-axis, is :

Accepted Answer

$\begin{gathered} {y}^{2}=4(\frac{1}{4})(x-2) \ y=m(x-2)+\frac{1}{4m}(-2,0) \ \Rightarrow 0=-4m+\frac{1}{4m}\Rightarrow 16{m}^{2}=1 \ \Rightarrow m=\pm \frac{1}{4} \ m=\frac{1}{4}\Rightarrow (6,2) \end{gathered}$

$\begin{gathered} \Rightarrow =\frac{1}{2}	imes (1)	imes 4+{\int }_{6}^{2}[(\frac{x+2}{4})-\sqrt{x-2}]dx \ \, =2+\frac{2}{3}=\frac{8}{3} \end{gathered}$

Solutions · 1