Slope = dy/dx = 2x = 2 at x=1. Tangent: y - 1 = 2(x - 1).

Equation of Tangent to Curve y = x² at Point (1,1)

Question

Find the equation of the tangent to the curve $y = x^2$ at the point $(1, 1)$ .

Solution — Step by Step

Differentiate $y = x^2$ with respect to $x$ .

\frac{dy}{dx} = 2x

This gives us the slope of the tangent at any point on the curve.

Substitute $x = 1$ into the derivative.

m = \left.\frac{dy}{dx}\right|_{x=1} = 2(1) = 2

So the tangent at $(1, 1)$ has slope 2.

We know the tangent passes through $(1, 1)$ with slope $2$ . Use:

y - y_1 = m(x - x_1)

y - 1 = 2(x - 1)

Expand and rearrange:

y - 1 = 2x - 2

y = 2x - 1

Or equivalently: $2x - y - 1 = 0$

The equation of the tangent is $y = 2x - 1$ .

Why This Works

The derivative $\frac{dy}{dx}$ at a point gives the instantaneous rate of change — which geometrically is the slope of the tangent line at that exact point. This is the entire bridge between differentiation and geometry.

Once we have the slope, the tangent is just a straight line. We already know one point it passes through (the point of tangency), so point-slope form finishes the job in one step.

In CBSE 12 and JEE Main, tangent questions always follow this three-step rhythm: differentiate → substitute → point-slope form. Internalize this flow and you’ll never get stuck.

Alternative Method — Using Standard Line Equation

For the parabola $y = x^2$ , there’s a direct result worth knowing.

The tangent at point $(a, a^2)$ on $y = x^2$ is:

y = 2ax - a^2

For our point $(1, 1)$ , put $a = 1$ :

y = 2(1)x - (1)^2 = 2x - 1

Same answer. This shortcut is useful when the question gives you a parameter instead of a numerical point — common in JEE Main objective questions.

Common Mistake

Many students write the slope as $\frac{dy}{dx} = 2x$ and then forget to substitute $x = 1$ . They use slope $= 2x$ directly in point-slope form, getting a mess like $y - 1 = 2x(x - 1)$ . That’s the equation of a curve, not a line.

Always substitute the $x$ -coordinate of the point before building the tangent equation.

A second trap: confusing the tangent with the normal. The normal is perpendicular to the tangent, so its slope is $-\frac{1}{m} = -\frac{1}{2}$ . CBSE 2024 had both tangent and normal parts in the same question — read carefully which one is asked.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Standard Line Equation

Common Mistake

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