Parabola
Shortest distance between parabola and circle
MJAT_TS4_P1
Grade 12
Question:
Consider a parabola $P: x^2=4ky$ and a circle $C: x^2+y^2-2\alpha x+6y+9=0$ (where $\alpha>0$). The line of shortest distance between them cuts the parabola $P$ at the end point of the latus rectum in the first quadrant. If $k=1$, then $\alpha=$
Step-by-Step Solution
Key Concept: For $k=1$: $P: x^2=4y$. End of latus rectum in first quadrant: $(2k, k)=(2,1)$. The shortest distance line passes through $(2,1)$ and is perpendicular to the parabola's tangent at that point. Tangent at $(2,1)$: $x\cdot 2=2(y+1)\Rightarrow x=y+1$, slope $1$. So the shortest distance line has slope $-1$.
$\alpha=\mathbf{6}$.
Correct Answer: 6