Let the Normal at the Point P on the Parabola y² = 6x Pass Through the Point (5, −8). If the Tangent at P Intersects Its Directrix at the Point Q, Then the Ordinate of Q Is

JEE Main 2022 — 26 June, Morning Shift. Previous Year Question.

Problem

Let the normal at the point $P$ on the parabola $y^2=6x$ pass through the point $(5,-8)$. If the tangent at $P$ to the parabola intersects its directrix at the point $Q$, then the ordinate of the point $Q$ is:

Key insight. Writing $P$ in parametric form turns the normal’s “passes through $(5,-8)$” condition into a single cubic equation in the parameter $t$ — and since JEE cubics are almost always built with a clean integer root, testing small values first (rather than reaching for the general cubic formula) finds it quickly.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Identify $a$ from $y^2=6x$ and write $P$ in parametric form $(at^2,2at)$. Write the equation of the normal at $P$ and substitute the given point $(5,-8)$ to get a cubic equation in $t$; solve it by testing simple values. With $t$ known, write the tangent at $P$ and find where it crosses the directrix $x=-a$.

Solution

Step 1 — Identify a and write P parametrically

$y^2=6x$ matches $y^2=4ax$ with $4a=6\implies a=\dfrac32$. Point $P=(at^2,2at)=\left(\dfrac32t^2,3t\right)$.

Step 2 — Find the tangent slope, and hence the normal slope

For $y^2=4ax$, the slope of the tangent at parameter $t$ is $\dfrac1t$, so the normal’s slope is $-t$.

Step 3 — Write the equation of the normal

$$y-2at = -t(x-at^2)$$

With $a=\dfrac32$: $y-3t = -t\left(x-\dfrac32t^2\right) \implies y = -tx+\dfrac32t^3+3t$

Step 4 — Substitute the point (5, −8)

$$-8 = -5t+\frac32t^3+3t = \frac32t^3-2t$$

Multiplying by $2$:

$$-16 = 3t^3-4t \implies 3t^3-4t+16=0$$

Step 5 — Solve the cubic by testing simple values

Testing $t=-2$: $3(-2)^3-4(-2)+16 = -24+8+16=0$ ✓

So $t=-2$.

Step 6 — Write the tangent at P using t = −2

For $y^2=4ax$, the tangent at parameter $t$ is $ty=x+at^2$, i.e. $y=\dfrac{x}{t}+at$.

With $a=\dfrac32$, $t=-2$:

$$y = \frac{x}{-2}+\frac32(-2) = -\frac{x}{2}-3$$

Step 7 — Find where the tangent meets the directrix

Directrix: $x=-a=-\dfrac32$.

$$y = -\frac{1}{2}\left(-\frac32\right)-3 = \frac34-3 = \frac{3-12}{4} = -\frac94$$

Answer

Common mistakes

• Trying to solve the cubic $3t^3-4t+16=0$ using the general cubic formula. JEE-style cubics arising from these normal-passes-through-a-point conditions are constructed to have a clean rational root — testing a few small integer or simple-fraction values first (as with $t=-2$ here) is dramatically faster.
• Using the wrong sign convention in the tangent formula. The tangent to $y^2=4ax$ at parameter $t$ is $ty=x+at^2$ — mixing this up with the normal’s formula, or forgetting the factor of $t$ on the left, gives an entirely different line that won’t correctly intersect the directrix at $Q$.

Practise next

• Let the normal at a point $P$ on the parabola $y^2=8x$ pass through $(0,24)$. If the tangent at $P$ meets the directrix at $Q$, find the ordinate of $Q$, using the same parametric-normal-then-tangent approach.