Parabola
Normal with given slope — latus rectum and chord ratio
MJAT_TS7_P2
Grade 12
Question:
A normal with slope $\frac{1}{6}$ is drawn from $(0,-\alpha)$ to the parabola $x^2=-4ay$ ($a>0$). Let $L$ be the line through $(0,-\alpha)$ parallel to the directrix, intersecting the parabola at $A$ and $B$. If $r:s=1:16$ where $r=4a$ (latus rectum length) and $s=|AB|^2$, then $24a$ equals:
Step-by-Step Solution
Key Concept: Parabola $x^2=-4ay$: vertex at origin, opens downward, directrix $y=a$. Normal at $(2at,-at^2)$: slope $=-1/t=1/6\Rightarrow t=-6$. Point on normal: $(−12a,−36a)$. So $\alpha=36a$. Line $y=-36a$ (parallel to directrix $y=a$). Intersect parabola: $x^2=-4a(-36a)=144a^2\Rightarrow x=\pm 12a$. $|AB|=24a\Rightarrow s=(24a)^2=576a^2$. $r=4a$. $r:s=4a:576a^2=1:(144a)$. Set $=1:16\Rightarrow 144a=16\Rightarrow a=1/9$.
$24a=\mathbf{12}$.
Correct Answer: 12