Differentiation (Part-38) | S N De

Tangent and Normal — Part 6 Solutions

In the previous article, we solved several Short Answer Type Questions. In this article, we continue discussing step-by-step solutions for S.N. Dey Mathematics (Class 12) for the chapter Tangent and Normal (Exercise 14).

Find the equations of the tangent and normal to each of the following curves at the specified points [7–11]:

7. $x = a\sec\theta, ~ y = b\tan\theta$ at the point $\theta.$

Solution:

$x = a\sec\theta \implies \frac{dx}{d\theta} = a\sec\theta\tan\theta$

$y = b\tan\theta \implies \frac{dy}{d\theta} = b\sec^2\theta$

$\therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b\sec^2\theta}{a\sec\theta\tan\theta} = \frac{b\sec\theta}{a\tan\theta}$

The equation of the tangent to the given curve at $\theta$ is:

$y - b\tan\theta = \frac{b\sec\theta}{a\tan\theta}(x - a\sec\theta)$

$\text{or, } ay\tan\theta - ab\tan^2\theta = bx\sec\theta - ab\sec^2\theta$

$\text{or, } ay\tan\theta - bx\sec\theta = -ab(\sec^2\theta - \tan^2\theta)$

$\text{or, } bx\sec\theta - ay\tan\theta = ab$

$\therefore \frac{x}{a}\sec\theta - \frac{y}{b}\tan\theta = 1$

The equation of the normal to the given curve at $\theta$ is:

$y - b\tan\theta = \left[-\frac{dx}{dy}\right]_{\theta}(x - a\sec\theta)$

$\text{or, } y - b\tan\theta = -\frac{a\tan\theta}{b\sec\theta}(x - a\sec\theta)$

$\text{or, } by\sec\theta - b^2\tan\theta\sec\theta = -ax\tan\theta + a^2\sec\theta\tan\theta$

$\text{or, } by\sec\theta + ax\tan\theta = (a^2 + b^2)\tan\theta\sec\theta$

$\therefore \frac{ax}{\sec\theta} + \frac{by}{\tan\theta} = a^2 + b^2$

8. $xy^2 = 18$ at the point $(2, 3).$

Solution:

$xy^2 = 18 \quad \longrightarrow (1)$

Differentiating (1) with respect to $x$:

$1 \cdot y^2 + x \cdot 2y \frac{dy}{dx} = 0$

$\text{or, } \frac{dy}{dx} = -\frac{y^2}{2xy} = -\frac{y}{2x}$

$\therefore \left[\frac{dy}{dx}\right]_{(2,3)} = -\frac{3}{2 \times 2} = -\frac{3}{4}$

$\left[-\frac{dx}{dy}\right]_{(2,3)} = \frac{4}{3}$

Equation of tangent at $(2, 3)$:

$y - 3 = -\frac{3}{4}(x - 2) \implies 4y - 12 = -3x + 6$

$\therefore \mathbf{3x + 4y = 18}$

Equation of normal at $(2, 3)$:

$y - 3 = \frac{4}{3}(x - 2) \implies 3y - 9 = 4x - 8$

$\therefore \mathbf{4x - 3y + 1 = 0}$

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9. $y(x-2)(x-3) + 7 = x$ at the point where the curve intersects the x-axis.

Solution:

$y(x-2)(x-3) = x - 7 \quad \longrightarrow (1)$

$\log[y(x-2)(x-3)] = \log(x-7) \quad \longrightarrow (2)$

Differentiating (2) with respect to $x$:

$\frac{1}{y}\frac{dy}{dx} + \frac{1}{x-2} + \frac{1}{x-3} = \frac{1}{x-7}$

$\frac{dy}{dx} = \frac{1}{(x-2)(x-3)} - \frac{x-7}{(x-2)^2(x-3)} - \frac{x-7}{(x-2)(x-3)^2}$

The curve intersects the x-axis where $y = 0 \implies x = 7$. Point of contact is $(7, 0)$.

$\left[\frac{dy}{dx}\right]_{(7,0)} = \frac{1}{(5)(4)} - 0 - 0 = \frac{1}{20}$

Equation of tangent at $(7, 0)$:

$y - 0 = \frac{1}{20}(x - 7) \implies \mathbf{x - 20y = 7}$

Equation of normal at $(7, 0)$:

$y - 0 = -20(x - 7) \implies \mathbf{20x + y = 140}$

10. $x = a(\theta - \sin\theta), ~ y = a(1 - \cos\theta)$ at $\theta = \pi.$

Solution:

$\frac{dx}{d\theta} = a(1 - \cos\theta), \quad \frac{dy}{d\theta} = a\sin\theta$

$\frac{dy}{dx} = \frac{\sin\theta}{1 - \cos\theta} \implies \left[\frac{dy}{dx}\right]_{\theta=\pi} = \frac{0}{1 - (-1)} = 0$

At $\theta = \pi$:

$x = a(\pi - \sin\pi) = a\pi, \quad y = a(1 - \cos\pi) = 2a$

Equation of tangent at $(a\pi, 2a)$:

$y - 2a = 0(x - a\pi) \implies \mathbf{y = 2a}$

Equation of normal at $(a\pi, 2a)$:

$x - a\pi = 0(y - 2a) \implies \mathbf{x = a\pi}$

11. $x^{2/3} + y^{2/3} = 2$ at $(1, 1).$

Solution:

$\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\sqrt[3]{\frac{y}{x}}$

$\left[\frac{dy}{dx}\right]_{(1,1)} = -1, \quad \left[-\frac{dx}{dy}\right]_{(1,1)} = 1$

Equation of tangent at $(1, 1)$:

$y - 1 = -1(x - 1) \implies \mathbf{x + y = 2}$

Equation of normal at $(1, 1)$:

$y - 1 = 1(x - 1) \implies \mathbf{x - y = 0}$

12. Show that the equation of the normal to the hyperbola $x = a\sec\theta, y = b\tan\theta$ at $(a\sec\theta, b\tan\theta)$ is $ax\cos\theta + by\cot\theta = a^2 + b^2.$

Solution:

$\frac{dx}{d\theta} = a\sec\theta\tan\theta, \quad \frac{dy}{d\theta} = b\sec^2\theta$

$\frac{dy}{dx} = \frac{b\sec\theta}{a\tan\theta} \implies -\frac{dx}{dy} = -\frac{a\tan\theta}{b\sec\theta}$

Equation of normal:

$y - b\tan\theta = -\frac{a\tan\theta}{b\sec\theta}(x - a\sec\theta)$

$by\sec\theta - b^2\tan\theta\sec\theta = -ax\tan\theta + a^2\tan\theta\sec\theta$

$ax\tan\theta + by\sec\theta = (a^2 + b^2)\sec\theta\tan\theta$

Dividing by $\sec\theta\tan\theta$: $\mathbf{ax\cos\theta + by\cot\theta = a^2 + b^2}$

13. Find the equation of the tangent to the circle $x^2 + y^2 = a^2$ at $(a\cos\theta, a\sin\theta).$ Hence show that $y = x + a\sqrt{2}$ touches the circle and find the point of contact.

Solution:

$2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$

$\left[\frac{dy}{dx}\right]_{(a\cos\theta, a\sin\theta)} = -\cot\theta$

Equation of tangent at $(a\cos\theta, a\sin\theta)$:

$y - a\sin\theta = -\frac{\cos\theta}{\sin\theta}(x - a\cos\theta)$

$\mathbf{x\cos\theta + y\sin\theta = a} \quad \longrightarrow (1)$

Given line $y = x + a\sqrt{2}$ has slope $m = 1$. Equating slopes:

$-\cot\theta = 1 \implies \theta = \frac{3\pi}{4}$

Substituting $\theta = \frac{3\pi}{4}$ into (1):

$x\left(-\frac{1}{\sqrt{2}}\right) + y\left(\frac{1}{\sqrt{2}}\right) = a \implies y = x + a\sqrt{2}$

Hence, the line touches the circle.

Point of contact: $\left(a\cos\frac{3\pi}{4}, a\sin\frac{3\pi}{4}\right) = \mathbf{\left(-\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)}$