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Tangent to the Curve at a Point - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Slope and Equation of Normal is considered one of the most asked concept.
65 Questions around this concept.

Solve by difficulty

Angle between the tangents to the curve $y=x^{2}-5x+6$ at the points $\left ( 2,0 \right )$ and $\left ( 3,0 \right )$ is

$\frac{\pi }{2}$

$\frac{\pi }{3}$

$\frac{\pi }{6}$

$\frac{\pi }{4}$

View Solution

The normal to the curve, x² + 2xy - 3y² = 0, at (1,1):

does not meet the curve again.

meets the curve again in the second quadrant.

meets the curve again in the third quadrant.

meets the curve again in the fourth quadrant.

View Solution

Concepts Covered - 0

Slope and Equation of Tangent

Tangent and Normal

Slope and Equation of Tangent

Let P(x_o, y_o) be a point on the continuous curve y = f(x), then the slope of the tangent to the curve at point P is

$\left ( \frac{dy}{dx} \right )_{\left (x_0,y_0 \right )} .$

$\Rightarrow \left(\frac{d y}{d x}\right)_{\left ( x_0,y_0 \right )}=\tan \theta=\text { slope of tangent at } P$

Where Ө is the angle which the tangent at P(x_o, y_o) makes with the positive direction of the x-axis as shown in the figure.

If the tangent is parallel to x-axis then Ө = 0^o.

$\\\Rightarrow\;\;\;\;\; \tan\theta=0\\\\\therefore\;\; \left (\frac{dy}{dx} \right )_{(x_0,y_0)}=0$

If the tangent is perpendicular to x-axis then Ө = 90^o

$\\\Rightarrow\;\;\;\;\; \tan\theta\rightarrow \infty\;\;\;\;\text{or}\;\;\;\;\cot\theta=0\\\\\therefore\;\; \left (\frac{dx}{dy} \right )_{(x_0,y_0)}=0$

Equation of Tangent

Let the equation of curve be y = f (x) and let point P (x₀, y₀) lies on this curve.

The slope of the tangent to the curve at a point P is

$\left ( \frac{dy}{dx} \right )_{\left (x_0,y_0 \right )} .$

Hence, the equation of the tangent at point P is

$(y-y_0)=\left ( \frac{dy}{dx} \right )_{(x_0,y_0)}\cdot(x-x_0)$

Tangent from External Point

If a point Q(a, b) does not lie on the curve y = f(x), then the equation of possible tangent to the curve y = f(x) (tangent passing through point Q (a, b)) can be found by first getting the point of contact P(x_o, y_o) on the curve.

$\\P\;(x_0,y_0)\;\text{lies on the curve y = f(x), then}\\\\\text{\;\;\;\;\;\;\;}y_0=f(x_0)\\\\\text{Also, slope of PQ is}\\\\\text{\;\;\;\;\;\;}\frac{y_0-b}{x_0-a}=\left ( \frac{dy}{dx} \right )_{(x_0,y_0)}$

By solving the above two equations we get point of contact point P.

Using P we can find the equation of tangent PQ

Slope and Equation of Normal

Slope and Equation of Normal

The normal to a curve at a given point say P(x_o, y_o) is a line perpendicular to the tangent at P and which passes through P.

So, let the continuous curve be y = f(x) and let the point P(x_o, y_o) lies on this curve.

Therefore the slope of normal to the curve at point P(x_o, y_o) is given by :

$\\\text{Slope of normal at P}=-\mathrm{\frac{1}{slope \;of\;tangent \;at\;P }}\\\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =-\frac{1}{\left ( \frac{dy}{dx} \right )_{(x_0,y_0)}}}\\\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =-\left (\frac{dx}{dy } \right )_{(x_0,y_0)}}$