Find the locus of mid-point of chord of parabola y²=4ax with given slope

Question

Find the locus of the mid-point of chords of the parabola $y^2 = 4ax$ that have slope $m$ .

(JEE Advanced 2022, similar pattern)

Solution — Step by Step

Let the mid-point of the chord be $P(h, k)$ . We need to find the relationship between $h$ and $k$ (the locus equation).

For any conic, the equation of the chord with mid-point $(h, k)$ is given by $T = S_1$ .

For the parabola $y^2 = 4ax$ :

$T$ : Replace $y^2$ with $yk$ , and $x$ with $\frac{x + h}{2}$ . So $T \equiv ky - 2a(x + h)$
$S_1$ : Substitute the mid-point into the parabola equation: $S_1 = k^2 - 4ah$

The chord equation: $ky - 2a(x + h) = k^2 - 4ah$

Simplifying: $ky - 2ax = k^2 - 2ah$

Rearrange: $ky = 2ax + k^2 - 2ah$

y = \frac{2a}{k}x + \frac{k^2 - 2ah}{k}

The slope of this chord is $\frac{2a}{k}$ .

We’re told the slope equals $m$ :

\frac{2a}{k} = m

k = \frac{2a}{m}

Replace $(h, k)$ with $(x, y)$ to get the locus:

\boxed{y = \frac{2a}{m}}

This is a horizontal straight line.

Why This Works

The $T = S_1$ method is a powerful shortcut for finding the equation of a chord when you know its mid-point. It works because the chord bisected at $(h, k)$ must be perpendicular to the line joining $(h, k)$ to… wait, that’s circles. For parabolas, it’s different.

The result says all chords of slope $m$ have their mid-points on the horizontal line $y = 2a/m$ . This makes geometric sense — as you slide a chord of fixed slope along the parabola, its mid-point traces a horizontal path. Steeper chords ( $m$ large) have mid-points closer to the axis ( $y$ small), while nearly horizontal chords ( $m$ small) have mid-points far from the axis.

Alternative Method — Parametric approach

Let the two endpoints of the chord be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ on the parabola.

Mid-point: $h = \frac{a(t_1^2 + t_2^2)}{2}$ , $k = \frac{2a(t_1 + t_2)}{2} = a(t_1 + t_2)$

Slope of chord: $\frac{2at_1 - 2at_2}{at_1^2 - at_2^2} = \frac{2}{t_1 + t_2} = m$

So $t_1 + t_2 = \frac{2}{m}$ , giving $k = a \cdot \frac{2}{m} = \frac{2a}{m}$ . Same result.

The $T = S_1$ formula is a universal tool for conics. For any conic $S = 0$ , the chord with mid-point $(h, k)$ has the equation $T(h, k) = S_1(h, k)$ . Learn this once and apply it to circles, ellipses, parabolas, and hyperbolas — it saves enormous time in JEE.

Common Mistake

Students often forget to replace $(h, k)$ with $(x, y)$ at the end, leaving the answer as $k = 2a/m$ . The question asks for the locus, which must be expressed in terms of $x$ and $y$ . Also, a common algebraic slip is writing $T$ incorrectly — for $y^2 = 4ax$ , the $T$ involves $2a(x + h)$ , not $4a(x + h)$ . Be careful with the factor of 2.

Question

Solution — Step by Step

Why This Works

Alternative Method — Parametric approach

Common Mistake

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