Analytic Geometry – Problem 10: Find the slope of

The directrix of parabola $E:\,y^2=4x$ meets the $x$-axis at point $E$. A line through the focus $F$ intersects the parabola at $A$, $B$. Points $C$, $D$ lie on the directrix $l$ satisfying $AC\perp l$ and $BD\perp l$. The circle with diameter $CD$ is tangent to line $AB$ at $F$. The quadrilateral $ABDC$ has area $64$. Find the slope of $AB$.

Since $AC\perp l$ and $BD\perp l$, by the parabola definition for $y^2=4x$, we have $AF=AC$ and $BF=BD$. Let the slope of $AB$ be $k$ (with $k\neq0$). Set $$AB:\ y=k(x-1).$$ Substitute into $y^2=4x$: $$k^2x^2-(2k^2+4)x+k^2=0.$$ Let the intersection abscissae be $x_1,x_2$. Then $$x_1+x_2=\frac{2k^2+4}{k^2},\qquad x_1x_2=1.$$

For $y^2=4x$, directrix is $x=-1$, so distance from $(x_i,y_i)$ to directrix is $x_i+1$. Hence $$AC=x_1+1,\quad BD=x_2+1,$$ thus $$AC+BD=x_1+x_2+2=\frac{2k^2+4}{k^2}+2=\frac{4(k^2+1)}{k^2}.$$

Also $$CD=|y_1-y_2|=\sqrt{(y_1+y_2)^2-4y_1y_2}.$$ Because $y_i=k(x_i-1)$, $$y_1+y_2=k(x_1+x_2)-2k=\frac{4}{k},\qquad y_1y_2=-4,$$ so $$CD=\sqrt{\frac{16}{k^2}+16}=\frac{4\sqrt{k^2+1}}{|k|}.$$

Quadrilateral $ABDC$ is a right trapezoid with parallel sides $AC,BD$ and height $CD$, therefore $$S=\frac12(AC+BD)\cdot CD=\frac12\cdot\frac{4(k^2+1)}{k^2}\cdot\frac{4\sqrt{k^2+1}}{|k|}=64.$$ Hence $$\frac{8(k^2+1)^{3/2}}{|k|^3}=64\quad\Rightarrow\quad\frac{(k^2+1)^{3/2}}{|k|^3}=8.$$ Let $u=k^2+1\ge1$. Then $$\left(\frac{u}{u-1}\right)^{3/2}=8\quad\Rightarrow\quad\frac{u}{u-1}=4\quad\Rightarrow\quad u=\frac43.$$ So $k^2=u-1=\frac13$, and $$k=\pm\frac{\sqrt3}{3}.$$

$$k_{AB}=\pm\dfrac{\sqrt{3}}{3}.$$