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

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

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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

$\frac{\pi }{2}$

$\frac{\pi }{3}$

$\frac{\pi }{6}$

$\frac{\pi }{4}$

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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.

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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 :

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