Tangents and normals

This topic is not in the rationalised NCERT chapter but is still asked in JEE Advanced.

The slope of the tangent to y = f(x) at (x₀, y₀) is f′(x₀). The tangent line is y − y₀ = f′(x₀)(x − x₀).

The normal is perpendicular to the tangent. When f′(x₀) ≠ 0 its slope is −1/f′(x₀), so its equation is y − y₀ = −(x − x₀)/f′(x₀).

Worked example: y = x² at (1, 1) has slope 2. The tangent is y = 2x − 1 and the normal is x + 2y = 3.

A zero slope gives a horizontal tangent and a vertical normal x = x₀. An infinite slope gives a vertical tangent x = x₀.

For a parametric curve, the slope is (dy/dt)/(dx/dt) at the parameter value of the point.