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

Question

Let the directrix of the parabola $P:{y}^{2}=8x$, cut $x$-axis at the point A . Let $B(\alpha ,\beta ),\alpha >1$, be a point on $P$ such that the slope of $AB$ is $3/5$. If $BC$ is a focal chord of $P$, then six times the area of $	riangleABC$ is :

Accepted Answer

For the parabola

${y}^{2}=8x$

we compare it with the standard form

${y}^{2}=4ax$

So,

$4a=8\, \implies \, a=2$

Hence:

focus $S=(2,0)$

directrix is

$x=-2$

Since the directrix cuts the $x$-axis at point $A$, we get

$A=(-2,0)$

Now let $B(\alpha ,\beta )$ be a point on the parabola such that the slope of $AB$ is $\frac{3}{5}$.

The slope of $AB$ is

$\frac{\beta -0}{\alpha -(-2)}=\frac{\beta }{\alpha +2}$

Given this is $\frac{3}{5}$, so

$\frac{\beta }{\alpha +2}=\frac{3}{5}$

Thus,

$5\beta =3(\alpha +2)$

Also, since $B$ lies on the parabola ${y}^{2}=8x$,

${\beta }^{2}=8\alpha $

Now substitute from (1). From (1),

$\beta =\frac{3(\alpha +2)}{5}$

Put into (2):

${(\frac{3(\alpha +2)}{5})}^{2}=8\alpha $

$\frac{9(\alpha +2{)}^{2}}{25}=8\alpha $

$9(\alpha +2{)}^{2}=200\alpha $

$9({\alpha }^{2}+4\alpha +4)=200\alpha $

$9{\alpha }^{2}+36\alpha +36=200\alpha $

$9{\alpha }^{2}-164\alpha +36=0$

Now solve:

$9{\alpha }^{2}-164\alpha +36=0$

Discriminant:

$D={164}^{2}-4\cdot 9\cdot 36=26896-1296=25600$

$\sqrt{D}=160$

So,

$\alpha =\frac{164\pm 160}{18}$

Hence,

$\alpha =\frac{324}{18}=18\, \, \alpha =\frac{4}{18}=\frac{2}{9}$

Given $\alpha >1$, we take

$\alpha =18$

Then from ${\beta }^{2}=8\alpha $,

${\beta }^{2}=144\, \implies \, \beta =\pm 12$

Since slope is $\frac{3}{5}>0$, $\beta $ must be positive. Therefore

$B=(18,12)$

Now $BC$ is a focal chord of the parabola.

For parabola ${y}^{2}=4ax$, a point in parametric form is

$(a{t}^{2},\, 2at)$

Here $a=2$, so point is

$(2{t}^{2},\, 4t)$

For point $B=(18,12)$,

$2{t}^{2}=18\, \implies \, {t}^{2}=9\, \implies \, t=3$

and

$4t=12\, \implies \, t=3$

So $B$ corresponds to parameter $t=3$.

If $BC$ is a focal chord, then the other endpoint corresponds to parameter $-\frac{1}{t}$.

Thus for $t=3$, point $C$ corresponds to parameter $-\frac{1}{3}$.

So,

$C=(2(\frac{1}{9}),\, 4(-\frac{1}{3}))=(\frac{2}{9},\, -\frac{4}{3})$

Now find area of $	riangleABC$.

Coordinates are:

$A=(-2,0),\, B=(18,12),\, C=(\frac{2}{9},-\frac{4}{3})$

Using determinant formula,

$=\frac{1}{2}|{x}_{1}({y}_{2}-{y}_{3})+{x}_{2}({y}_{3}-{y}_{1})+{x}_{3}({y}_{1}-{y}_{2})|$

So,

$=\frac{1}{2}[-2(12+\frac{4}{3})+18(-\frac{4}{3})+\frac{2}{9}(0-12)]|$

Now simplify:

$12+\frac{4}{3}=\frac{40}{3}$

So,

$-2\cdot \frac{40}{3}=-\frac{80}{3}$

Also,

$18\cdot (-\frac{4}{3})=-24$

and

$\frac{2}{9}(-12)=-\frac{24}{9}=-\frac{8}{3}$

Therefore,

$=\frac{1}{2}|-\frac{80}{3}-24-\frac{8}{3}|$

$=\frac{1}{2}|-\frac{88}{3}-24|=\frac{1}{2}|-\frac{88}{3}-\frac{72}{3}|$

$=\frac{1}{2}|-\frac{160}{3}|=\frac{80}{3}$

Hence,

$6	imes =6\cdot \frac{80}{3}=160$

Therefore, the required answer is

$160$

So, the correct option is B.

Solutions · 1