Let P:y^2=4ax, a0, be a parabola with focus S.

Question

Let $P:y^2=4ax$, $a>0$, be a parabola with focus $S$. Let the tangents to $P$ that make an angle of $\dfrac\pi4$ with the line $y=3x+5$ touch $P$ at $A$ and $B$. Then the value of $a$ for which $A,B,S$ are collinear is:

Accepted Answer

$\tan45^\circ=\left|\dfrac{m-3}{1+3m}\right|=1\Rightarrow m=-2$ or $m=\dfrac12$.

Points of contact: $A=\left(\dfrac a4,-a\right)$ (for $m=-2$), $B=(4a,4a)$ (for $m=\tfrac12$). With $S=(a,0)$:

$$\text{Area}(ABS)=\frac12\left|\frac a4(4a-0)+a(4a-a)+1(0-4a^2)\right|=\frac12\left|a^2+3a^2-4a^2\right|=0$$

for every $a$, so $A,B,S$ are collinear for any $a>0$.

Solutions · 1