Derive tangent equation ty = x + at² using parametric form.

Equation of Tangent to Parabola y² = 4ax at Point (at², 2at)

Question

Find the equation of the tangent to the parabola $y^2 = 4ax$ at the point $P(at^2, 2at)$ , where $t$ is the parameter.

Solution — Step by Step

We need the slope of the tangent at our point. Differentiating $y^2 = 4ax$ with respect to $x$ :

2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}

At $P(at^2, 2at)$ , the $y$ -coordinate is $2at$ . Plug it in:

\frac{dy}{dx}\bigg|_P = \frac{2a}{2at} = \frac{1}{t}

So the slope of the tangent at parameter $t$ is $\frac{1}{t}$ .

Tangent through $(at^2, 2at)$ with slope $\frac{1}{t}$ :

y - 2at = \frac{1}{t}(x - at^2)

ty - 2at^2 = x - at^2

ty = x + at^2

\boxed{ty = x + at^2}

This is the standard result you must memorise for JEE. The parameter $t$ appears in both the slope and the intercept — that’s the elegance of parametric form.

Why This Works

The parametric point $(at^2, 2at)$ lies on $y^2 = 4ax$ by construction — substitute and verify: $(2at)^2 = 4at^2 \cdot a$ . We’re not guessing a point; we’re guaranteed it’s on the curve for every real $t$ .

The implicit differentiation approach gives us slope as a function of $y$ , not $x$ . This is actually cleaner for parabolas because $y$ appears linearly in the derivative, whereas $x$ would require a square root. Substituting $y = 2at$ directly gives the slope $\frac{1}{t}$ without any messy algebra.

The result $ty = x + at^2$ is a linear equation — that confirms it’s genuinely a line, not a curve. The key insight is that $t$ acts as a single-parameter description of every tangent to this parabola.

Alternative Method — Using the Condition for Tangency

We can derive the same result by assuming the tangent has the form $y = mx + c$ and imposing the condition that it touches the parabola.

Substitute $y = mx + c$ into $y^2 = 4ax$ :

(mx + c)^2 = 4ax

m^2x^2 + (2mc - 4a)x + c^2 = 0

For tangency, the discriminant must be zero (equal roots):

(2mc - 4a)^2 - 4m^2c^2 = 0

4m^2c^2 - 16amc + 16a^2 - 4m^2c^2 = 0

-16amc + 16a^2 = 0 \implies c = \frac{a}{m}

So the tangent is $y = mx + \frac{a}{m}$ . Now, at our point $P(at^2, 2at)$ the slope $m = \frac{1}{t}$ , giving $c = at$ . Substituting back:

y = \frac{x}{t} + at \implies ty = x + at^2 \checkmark

The formula $y = mx + \frac{a}{m}$ is the tangent to $y^2 = 4ax$ in slope form. It’s directly useful when the slope $m$ is given in the problem, instead of the parameter $t$ . JEE Main 2023 had a variant where they gave the slope — this form saves a full step.

Common Mistake

The most common error: students differentiate $y^2 = 4ax$ and write $\frac{dy}{dx} = \frac{2a}{y}$ , then substitute $x = at^2$ instead of $y = 2at$ . The derivative is expressed in terms of $y$ , so you must substitute the $y$ -coordinate. Substituting $x$ here gives a completely wrong slope and loses all the marks for the remaining steps.

A second trap — forgetting to verify the point lies on the curve before using it. If you’re ever given a general parametric point in a JEE problem, always do a quick sanity check: does it satisfy the curve equation? For $(at^2, 2at)$ on $y^2 = 4ax$ : $(2at)^2 = 4a^2t^2 = 4a \cdot at^2$ . Yes, it does.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Condition for Tangency

Common Mistake

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