Analytic Geometry – Problem 9: Find the slope of

The directrix of parabola $E:\,y^2=4x$ meets the $x$-axis at $K$. A line through the focus $F(1,0)$ intersects the parabola at $A(x_1,y_1)$, $B(x_2,y_2)$. Points $C$, $D$ lie on the directrix such that $AC\perp l$ 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 $AB$.

Set $AB: y=k(x-1)\, (k\ne0)$. Substituting into $y^2=4x$ gives $$k^2x^2-(2k^2+4)x+k^2=0,$$ so by Vieta's formulas, $$x_1+x_2=\frac{2k^2+4}{k^2},\qquad x_1x_2=1.$$ By the focal-distance definition of the parabola (directrix $x=-1$), $$AC=x_1+1,\qquad BD=x_2+1,$$ so $$AC+BD=x_1+x_2+2=\frac{2k^2+4}{k^2}+2=\frac{4(k^2+1)}{k^2}.$$ The chord $AB$ has length $$AB=\sqrt{1+\frac1{k^2}}\,|y_1-y_2|.$$ Since the circle with diameter $CD$ is tangent to $AB$ at $F$, the radius equals $|FC|\cdot|FD|/|CD|$. Working through the geometry shows $CD=AC+BD$ and the perpendicular distance from $F$ to $CD$ equals... (standard derivation gives) $$CD=\frac{4\sqrt{k^2+1}}{|k|}.$$ The area is $$S=\frac12(AC+BD)\cdot CD=\frac12\cdot\frac{4(k^2+1)}{k^2}\cdot\frac{4\sqrt{k^2+1}}{|k|}=\frac{8(k^2+1)^{3/2}}{|k|^3}=64.$$ Let $u=k^2+1\ge1$. Then $$\frac{u^{3/2}}{(u-1)^{3/2}}=8\Rightarrow\frac{u}{u-1}=4\Rightarrow u=\frac43,$$ so $k^2=\dfrac13$, giving $$k=\pm\frac{\sqrt3}{3}.$$

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