Parabola
Tangent from external point — orthocenter and foot of perpendicular
MJAT_TS6_P2
Grade 12
Question:
A line parallel to axis of $y^2=4x$ through $P(\alpha-5,\alpha)$ meets parabola at $Q$. Tangent at $Q$: $x-2y+4=0$. $T$ on tangent; $M,N$ feet of perpendiculars from $T$ on $SQ$ and directrix. Which is/are correct?
A) $\alpha = -4$
B) If $PA,PB$ tangents at $A,B$, distance of orthocentre of $\triangle PAB$ from origin is $\sqrt{17}$
C) Area of $\triangle PAB = \dfrac{20}{5}$
D) If $QM=3$ then $TN$ can be $2$
Step-by-Step Solution
Key Concept: Tangent $x-2y+4=0$: at $(t^2,2t)$: tangent is $ty=x+t^2$, so $t=-2$, $Q=(4,4)$. Line $y=\alpha$ (parallel to axis) meets at $y=4\Rightarrow\alpha=4$ (not $-4$, A ✗). $P=(-1,4)$ on directrix.
B ✓, C ✓, D ✓. Answer: B, C, D.
Correct Answer: BCD