JEE Main 2019 — 9 April, Morning Shift. Previous Year Question.
Problem
If one end of a focal chord of the parabola $y^2=16x$ is at $(1,4)$, then the length of this focal chord is:
Key insight. A focal chord’s other endpoint is found the same way any line meets a curve — write the line through the known point and the focus, substitute it into the parabola’s equation, and the resulting quadratic hands you both endpoints.
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Approach
Identify the focus of $y^2=16x$, then write the equation of the line through the given point $(1,4)$ and the focus — this line is the focal chord itself. Substitute that line into the parabola’s equation to get a quadratic in $x$; its two roots are the $x$-coordinates of the chord’s two endpoints. Finally, apply the distance formula between those two endpoints to get the chord length.
Solution
Step 1 — Identify a and the focus
Writing $y^2=16x$ in the standard form $y^2=4ax$ gives $4a=16$, so $a=4$. The focus of this parabola is therefore at $(a,0)=(4,0)$.
Step 2 — Equation of the focal chord
The chord passes through the given point $(1,4)$ and the focus $(4,0)$. Its slope is
$$m = \frac{0-4}{4-1} = -\frac43$$
so the line is
$$y-0 = -\frac43(x-4) \implies y = -\frac43(x-4)$$
Step 3 — Substitute into the parabola’s equation
Substituting this expression for $y$ into $y^2=16x$:
$$\left[-\frac43(x-4)\right]^2 = 16x \implies \frac{16}9(x-4)^2 = 16x$$
Dividing both sides by 16:
$$(x-4)^2 = 9x$$
Step 4 — Solve the resulting quadratic
Expanding and rearranging:
$$x^2-8x+16 = 9x \implies x^2-17x+16=0$$
Factoring by splitting the middle term:
$$x^2-x-16x+16=0 \implies x(x-1)-16(x-1)=0 \implies (x-1)(x-16)=0$$
So $x=1$ or $x=16$. Since $x=1$ is the already-known endpoint, the other endpoint has $x=16$.
Step 5 — Find the y-coordinate of the second endpoint
Substituting $x=16$ into the line equation from Step 2:
$$y = -\frac43(16-4) = -\frac43(12) = -16$$
So the other endpoint is $(16,-16)$.
Step 6 — Apply the distance formula
The length of the focal chord is the distance between $(1,4)$ and $(16,-16)$:
$$\text{Length} = \sqrt{(16-1)^2+(-16-4)^2} = \sqrt{15^2+(-20)^2} = \sqrt{225+400} = \sqrt{625} = 25$$
Answer
$$25$$
Common mistakes
- Sign errors when substituting the negative slope. With $m=-\frac43$, it’s easy to drop the negative sign somewhere in the substitution, which changes the resulting quadratic and gives the wrong second endpoint.
- Stopping after finding x=16 without computing y. The $x$-coordinate alone isn’t enough — the second endpoint’s $y$-coordinate must come from substituting back into the line (or the parabola) equation, and skipping this step makes the distance formula impossible to apply correctly.
- Forgetting to check that (1,4) actually lies on the parabola first. It’s worth confirming $4^2=16(1)$ holds before proceeding — building the whole solution on a point that isn’t even on the curve would silently invalidate everything that follows.
Practise next
- One end of a focal chord of the parabola $y^2=12x$ is at $(3,6)$. Find the length of the chord using the same substitution approach.
Show answer
$12$. Here $4a=12$, so $a=3$ and points are $(3t^2,6t)$. The given end $(3,6)$ has $t=1$.
The other end of a focal chord has parameter $-\tfrac1t=-1$, giving $(3,-6)$.
Both ends have $x=3$, which is the focus’s $x$-coordinate — this chord is the latus rectum, and its length is $4a=12$.
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