Differentiation (Part-38) | S N De

Tangent and Normal | Part-5

In the previous article, we solved several VSA type questions. In the following article, we discuss and solve step-by-step Very Short Answer Type Questions from S.N. Dey Mathematics (Class 12), Chapter: Tangent and Normal (Ex-14).

Question 1(i)

Find the points on the hyperbola $2x^2 - 3y^2 = 6$ at which the slope of the tangent line is $(-1)$.

Solution

$$2x^2 - 3y^2 = 6 \quad \longrightarrow(1)$$

Differentiating (1) w.r.t. $x$, we get:

$$4x - 6y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{4x}{6y} = \frac{2x}{3y} \quad \longrightarrow(2)$$

Let $(h,k)$ be the point on the hyperbola (1) at which the slope of the tangent line is $(-1)$.

$$\left[\frac{dy}{dx}\right]_{(h,k)} = -1 \implies \frac{2h}{3k} = -1 \quad \text{[By (2)]} \implies h = -\frac{3}{2}k$$

Since the point $(h,k)$ lies on the hyperbola (1):

$$2h^2 - 3k^2 = 6$$ $$2 \left(-\frac{3}{2}k\right)^2 - 3k^2 = 6 \implies 2 \times \frac{9k^2}{4} - 3k^2 = 6$$ $$\frac{9k^2}{2} - 3k^2 = 6 \implies 9k^2 - 6k^2 = 12 \implies 3k^2 = 12 \implies k^2 = 4 \implies k = \pm 2$$

For $k = 2 \implies h = -\frac{3}{2} \times 2 = -3$.

For $k = -2 \implies h = -\frac{3}{2} \times (-2) = 3$.

Hence, the required points are $(-3,2)$ and $(3,-2)$. (Ans)

Question 1(ii)

The slope of the tangent line to the curve $x^3 - x^2 - 2x + y - 4 = 0$ at some point on it is $1$. Find the coordinates of such point or points.

Solution

$$x^3 - x^2 - 2x + y - 4 = 0 \implies y = 4 + 2x + x^2 - x^3 \quad \longrightarrow(1)$$

Differentiating (1) w.r.t. $x$, we get:

$$\frac{dy}{dx} = 0 + 2 + 2x - 3x^2 = 2 + 2x - 3x^2 \quad \longrightarrow(2)$$

Let the slope of the tangent line to the curve (1) at the point $(h,k)$ be $1$.

$$\left[\frac{dy}{dx}\right]_{(h,k)} = 1 \implies 2 + 2h - 3h^2 = 1 \quad \text{[By (2)]}$$ $$3h^2 - 2h - 1 = 0$$ $$h = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \times 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6}$$ $$\implies h = \frac{6}{6} = 1 \quad \text{or} \quad h = \frac{-2}{6} = -\frac{1}{3}$$

Since $(h,k)$ lies on (1), $k = 4 + 2h + h^2 - h^3$.

For $h = 1 \implies k = 4 + 2(1) + 1^2 - 1^3 = 6$.

For $h = -\frac{1}{3}$:

$$k = 4 + 2\left(-\frac{1}{3}\right) + \left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right)^3 = 4 - \frac{2}{3} + \frac{1}{9} + \frac{1}{27}$$ $$k = 4 + \frac{-18 + 3 + 1}{27} = 4 - \frac{14}{27} = 3\frac{13}{27}$$

Hence, the required points are $(1,6)$ and $\left(-\frac{1}{3}, 3\frac{13}{27}\right)$. (Ans)

Question 1(iii)

The slope of the normal to the parabola $3y^2 + 4y + 2 = x$ at a point on it is $8$. Find the coordinates of the point.

Solution

$$3y^2 + 4y + 2 = x \quad \longrightarrow(1)$$

Differentiating (1) w.r.t. $x$, we get:

$$\frac{d}{dx}(3y^2 + 4y + 2) = \frac{d}{dx}(x) \implies 6y\frac{dy}{dx} + 4\frac{dy}{dx} = 1$$ $$(6y + 4)\frac{dy}{dx} = 1 \implies \frac{dx}{dy} = 6y + 4 \implies -\frac{dx}{dy} = -6y - 4$$

Let $(h,k)$ be the point on the parabola (1) at which the slope of the normal is $8$.

$$\left[-\frac{dx}{dy}\right]_{(h,k)} = 8 \implies -6k - 4 = 8 \implies -6k = 12 \implies k = -2$$

Since the point $(h,k)$ lies on the parabola (1), $3k^2 + 4k + 2 = h$.

For $k = -2$:

$$h = 3(-2)^2 + 4(-2) + 2 = 12 - 8 + 2 = 6$$

Hence, the coordinates of the required point are $(6,-2)$. (Ans)

Question 2

Find a point on the parabola $y = x^2 - 6x + 9$, where the tangent is parallel to the line joining the points $(4,1)$ and $(3,0)$.

Solution

$$y = x^2 - 6x + 9 \implies \frac{dy}{dx} = 2x - 6$$

Let the required point be $(h,k)$.

The slope of the straight line joining the points $(4,1)$ and $(3,0)$ is:

$$m = \frac{1 - 0}{4 - 3} = 1$$

By the given condition:

$$\left[\frac{dy}{dx}\right]_{(h,k)} = 1 \implies 2h - 6 = 1 \implies 2h = 7 \implies h = \frac{7}{2}$$

Since the point $(h,k)$ lies on the given parabola, $k = h^2 - 6h + 9$.

For $h = \frac{7}{2}$:

$$k = \left(\frac{7}{2}\right)^2 - 6\left(\frac{7}{2}\right) + 9 = \frac{49}{4} - 21 + 9 = \frac{49}{4} - 12 = \frac{1}{4}$$

Hence, the required point is $\left(\frac{7}{2}, \frac{1}{4}\right)$. (Ans)

Find the equations of the tangent and normal to each of the following curves at the specified points [3–11]:

Question 3

Find the equations of the tangent and normal to $y = x^2 + 4x + 1$ at $x = 3$.

Solution

$$y = x^2 + 4x + 1 \quad \longrightarrow(1)$$ $$\frac{dy}{dx} = 2x + 4 \implies \left[\frac{dy}{dx}\right]_{x=3} = 2(3) + 4 = 10$$

For $x = 3 \implies y = 3^2 + 4(3) + 1 = 9 + 12 + 1 = 22$. Thus, the point is $(3,22)$.

Equation of Tangent:

$$y - 22 = 10(x - 3) \implies 10x - y = 8 \quad \mathbf{\color{#059669}{(Ans)}}$$

Equation of Normal:

$$y - 22 = \left[-\frac{dx}{dy}\right]_{x=3}(x - 3) \implies y - 22 = -\frac{1}{10}(x - 3)$$ $$10(y - 22) = -(x - 3) \implies 10y - 220 = -x + 3 \implies x + 10y = 223 \quad \mathbf{\color{#059669}{(Ans)}}$$

Question 4

Find the equations of the tangent and normal to $y^2 = 4ax$ at the ends of its latus rectum.

Solution

The coordinates of the ends of the latus rectum for $y^2 = 4ax \quad \longrightarrow(1)$ are $(a, \pm 2a)$.

Differentiating (1) w.r.t. $x$:

$$2y\frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}$$

1. At point $(a, 2a)$:

$$\left[\frac{dy}{dx}\right]_{(a,2a)} = \frac{2a}{2a} = 1$$

Equation of Tangent: $y - 2a = 1 \cdot (x - a) \implies x - y + a = 0$

Equation of Normal: $y - 2a = -1 \cdot (x - a) \implies x + y = 3a$

2. At point $(a, -2a)$:

$$\left[\frac{dy}{dx}\right]_{(a,-2a)} = \frac{2a}{-2a} = -1$$

Equation of Tangent: $y - (-2a) = -1 \cdot (x - a) \implies x + y + a = 0$

Equation of Normal: $y - (-2a) = 1 \cdot (x - a) \implies x - y = 3a$

Tangents: $x - y + a = 0$ and $x + y + a = 0$ (Ans)
Normals: $x + y = 3a$ and $x - y = 3a$ (Ans)

Question 5

Find the equations of the tangent and normal to $y = x^3 - 3x$ at the point $(2,2)$.

Solution

$$y = x^3 - 3x \quad \longrightarrow(1)$$ $$\frac{dy}{dx} = 3x^2 - 3 \implies \left[\frac{dy}{dx}\right]_{(2,2)} = 3(2)^2 - 3 = 12 - 3 = 9$$ $$\left[-\frac{dx}{dy}\right]_{(2,2)} = -\frac{1}{9}$$

Equation of Tangent:

$$y - 2 = 9(x - 2) \implies 9x - y = 16 \quad \mathbf{\color{#059669}{(Ans)}}$$

Equation of Normal:

$$y - 2 = -\frac{1}{9}(x - 2) \implies 9(y - 2) = -(x - 2) \implies x + 9y = 20 \quad \mathbf{\color{#059669}{(Ans)}}$$

Question 6

Find the equations of the tangent and normal to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(a\cos\theta, b\sin\theta)$.

Solution

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies b^2 x^2 + a^2 y^2 = a^2 b^2 \quad \longrightarrow(1)$$

Differentiating (1) w.r.t. $x$:

$$b^2(2x) + a^2(2y)\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$$ $$\left[\frac{dy}{dx}\right]_{(a\cos\theta, b\sin\theta)} = -\frac{b^2 \cdot a\cos\theta}{a^2 \cdot b\sin\theta} = -\frac{b}{a} \cdot \frac{\cos\theta}{\sin\theta}$$

Equation of Tangent:

$$y - b\sin\theta = -\frac{b}{a} \cdot \frac{\cos\theta}{\sin\theta}(x - a\cos\theta)$$ $$ay\sin\theta - ab\sin^2\theta = -bx\cos\theta + ab\cos^2\theta$$ $$bx\cos\theta + ay\sin\theta = ab(\sin^2\theta + \cos^2\theta) = ab$$ $$\text{Dividing by } ab \implies \mathbf{\frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1} \quad \mathbf{\color{#059669}{(Ans)}}$$

Equation of Normal:

$$\left[-\frac{dx}{dy}\right]_{(a\cos\theta, b\sin\theta)} = \frac{a}{b} \cdot \frac{\sin\theta}{\cos\theta}$$ $$y - b\sin\theta = \frac{a}{b} \cdot \frac{\sin\theta}{\cos\theta}(x - a\cos\theta)$$ $$by\cos\theta - b^2\sin\theta\cos\theta = ax\sin\theta - a^2\sin\theta\cos\theta$$ $$ax\sin\theta - by\cos\theta = (a^2 - b^2)\sin\theta\cos\theta$$ $$\text{Dividing by } \sin\theta\cos\theta \implies \mathbf{\frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2} \quad \mathbf{\color{#059669}{(Ans)}}$$