Parabola — Tangents & Focal Distance
PYP_JEE_ADV_2026_P2
Grade None
Question:
Let $T$ be the tangent to the parabola $y^2=16x$ at the point $(64,32)$. Let $L$ be the tangent to the same parabola at another point $(x_1,y_1)$ on the parabola. If $L$ and $T$ are perpendicular to each other, then the distance between the point $(x_1,y_1)$ and the focus of the parabola, is
$\dfrac{15}{4}$
$4$
$\dfrac{17}{4}$
$5$
Step-by-Step Solution
Key Concept: For parabola $y^2=4ax$, the tangent at parameter $t$ has slope $1/t$. Perpendicular tangents satisfy $t_1t_2=-1$. Focal distance $= x_1+a$ (using focus-directrix property).
**Step 1: Find slope of T**
$y^2=16x \Rightarrow 4a=16, a=4$. Focus at $(4,0)$. At $(64,32)$: parametrically $y=8t \Rightarrow t=4$. Slope of tangent $T=1/t=1/4$.
**Step 2: Find $(x_1,y_1)$**
Tangent $L$ at parameter $t_1$ has slope $1/t_1$. $L\perp T$: $\frac{1}{t_1}\cdot\frac{1}{4}=-1 \Rightarrow t_1=-\frac{1}{4}$. So $(x_1,y_1)=(at_1^2,2at_1)=(4\cdot\frac{1}{16},8\cdot(-\frac{1}{4}))=(\frac{1}{4},-2)$.
**Step 3: Focal distance**
Distance from $(\frac{1}{4},-2)$ to focus $(4,0)$: $\sqrt{(4-\frac{1}{4})^2+4}=\sqrt{\frac{225}{16}+\frac{64}{16}}=\sqrt{\frac{289}{16}}=\dfrac{17}{4}$.
Correct Answer: C