Find the 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 $(at^2, 2at)$ .

(JEE Main 2022, similar pattern)

Solution — Step by Step

Differentiate $y^2 = 4ax$ implicitly:

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

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

At the point $(at^2, 2at)$ : slope $= \frac{2a}{2at} = \frac{1}{t}$ .

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

Multiply through by $t$ :

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

\boxed{ty = x + at^2}

The point $(at^2, 2at)$ should satisfy this equation: $t(2at) = at^2 + at^2$ , i.e., $2at^2 = 2at^2$ ✓

Also, this tangent has $y$ -intercept at $x = -at^2$ (set $y = 0$ : $0 = x + at^2 \Rightarrow x = -at^2$ ), which is the correct sub-tangent geometry for a parabola.

Why This Works

The parametric point $(at^2, 2at)$ lies on $y^2 = 4ax$ for every value of $t$ (verify: $(2at)^2 = 4a^2t^2 = 4a \cdot at^2$ ✓). The tangent equation $ty = x + at^2$ is one of the most important results in conic sections — it appears in nearly every JEE paper.

The slope $1/t$ tells us something useful: as $t \to 0$ (point near vertex), the tangent becomes vertical; as $t \to \infty$ (point far from vertex), the tangent becomes nearly horizontal. This matches the visual shape of the parabola.

Alternative Method — Using the tangent formula directly

For the parabola $y^2 = 4ax$ , the tangent at point $(x_1, y_1)$ is:

yy_1 = 2a(x + x_1)

Substitute $x_1 = at^2$ , $y_1 = 2at$ :

y(2at) = 2a(x + at^2)

2aty = 2a(x + at^2)

Divide by $2a$ : $ty = x + at^2$ . Same result.

Memorise the tangent at $(at^2, 2at)$ as $ty = x + at^2$ . For slope form, the tangent with slope $m$ is $y = mx + a/m$ . For the tangent at $(x_1, y_1)$ , use $yy_1 = 2a(x + x_1)$ . Having all three forms ready saves time in different question types.

Common Mistake

A frequent error: using $\frac{dy}{dx} = \frac{2a}{x}$ instead of $\frac{2a}{y}$ . This happens when students divide $4a$ by $2x$ instead of $2y$ during implicit differentiation. Remember — when you differentiate $y^2$ , you get $2y \cdot \frac{dy}{dx}$ , not $2x$ . The variable being differentiated is $y$ , and it appears on the left side.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the tangent formula directly

Common Mistake

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