# JEE Main & Advanced Mathematics Applications of Derivatives Slope of the Tangent and Normal

## Slope of the Tangent and Normal

Category : JEE Main & Advanced

(1) Slope of the tangent : If a tangent is drawn to the curve $y=f(x)$ at a point $P({{x}_{1}},\,{{y}_{1}})$ and this tangent makes an angle $\psi$ with positive x-direction then, ${{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=\tan \psi =$ Slope of the tangent.

·               If the tangent is parallel to x-axis, $\psi =0\Rightarrow {{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=0$

·               If the tangent is perpendicular to x-axis, $\psi =\frac{\pi }{2}\Rightarrow {{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}\to \,\,\,\infty$

(2) Slope of the normal : The normal to a curve at a point $P({{x}_{1}},\,{{y}_{1}})$ is a line perpendicular to the tangent at P and passing through P.  Slope of the normal $=\frac{-1}{\text{Slope of tangent }}=\frac{-1}{{{\left( \frac{dy}{dx} \right)}_{P({{x}_{1}},\,{{y}_{1}})}}}=-{{\left( \frac{dx}{dy} \right)}_{P({{x}_{1}},\,{{y}_{1}})}}$.

·         If the normal is parallel to x-axis,  $-{{\left( \frac{dx}{dy} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=0$ or $\frac{b}{a}=\frac{c}{b}$.

·         If the normal is perpendicular to x-axis (or parallel to y-axis),  $-{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}=0$.

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