Let the directrix of the parabola P: y^2=8x cut the x-axis at the point A.

Question

Let the directrix of the parabola $P: y^2=8x$ cut the $x$-axis at the point $A$. Let $B(\alpha,\beta)$, $\alpha>1$, be a point on $P$ such that the slope of $AB$ is $\dfrac35$. If $BC$ is a focal chord of $P$, then six times the area of $	riangle ABC$ is:

Accepted Answer

Here $4a=8\Rightarrow a=2$; directrix $x=-2$ meets the $x$-axis at $A=(-2,0)$.

Slope condition: $\dfrac{\beta}{\alpha+2}=\dfrac35\Rightarrow 5\beta=3(\alpha+2)$. Also $\beta^2=8\alpha$:

$$9(\alpha+2)^2=200\alpha \;\Rightarrow\; 9\alpha^2-164\alpha+36=0 \;\Rightarrow\; \alpha=18 \text{ (taking }\alpha>1).$$

Then $\beta^2=144$, and since $\beta$ must be positive (slope $3/5>0$), $\beta=12$, so $B=(18,12)$.

$B$ corresponds to parameter $t=3$ (since $(2t^2,4t)=(18,12)$). For focal chord $BC$, $C$ corresponds to $-1/t=-1/3$: $C=\left(\dfrac29,-\dfrac43\right)$.

Area of $\triangle ABC$ with $A=(-2,0),\ B=(18,12),\ C=\left(\dfrac29,-\dfrac43\right)$:

$$\text{Area}=\frac12\left|-2\left(12+\frac43\right)+18\left(-\frac43\right)+\frac29(0-12)\right|=\frac12\left|-\frac{80}{3}-24-\frac83\right|=\frac{80}{3}.$$

$$6\times\text{Area}=6\times\frac{80}{3}=160.$$

Solutions · 1