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

Question

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

Accepted Answer

$$\begin{aligned}
& y^2=4\left(\frac{1}{4}\right)(x-2) \
& y=m(x-2)+\frac{1}{4 m} \text { passes through }(-2,0) \
& \Rightarrow 0=-4 m+\frac{1}{4 m} \Rightarrow 16 m^2=1 \
& \Rightarrow m= \pm \frac{1}{4} \
& m=\frac{1}{4} \text { in first quadrant } \Rightarrow \text { contact point }(6,2)
\end{aligned}$$

$$
\begin{aligned}
& \Rightarrow \text { Area }=\frac{1}{2} \times(1) \times 4+\int_2^6\left[\left(\frac{x+2}{4}\right)-\sqrt{x-2}\right] d x \
& \quad=2+\frac{2}{3}=\frac{8}{3}
\end{aligned}$$

Solutions · 1