Find the equation of the normal at the specified point to each of the following curves [16–26]:
Find the equation of the normal to $y^2 = 4ax$ at $(0,0)$.
Solution
$$y^2 = 4ax \quad \longrightarrow(1)$$
Differentiating (1) w.r.t. $x$, we get:
$$2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}$$ $$\therefore -\frac{dx}{dy} = -\frac{y}{2a}$$ $$\text{So,} \quad \left[-\frac{dx}{dy}\right]_{(0,0)} = 0$$Hence, the equation of normal to the parabola (1) at $(0,0)$ is:
$$y - 0 = \left[-\frac{dx}{dy}\right]_{(0,0)}(x - 0)$$ $$\text{or,} \quad y = 0 \cdot x \implies y = 0 \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x,y)$.
Solution
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \implies b^2x^2 + a^2y^2 = a^2b^2 \quad \longrightarrow(1)$$
Differentiating (1) w.r.t. $x$, we get:
$$b^2 \cdot 2x + a^2 \cdot 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b^2x}{a^2y}$$ $$\therefore \left[-\frac{dx}{dy}\right]_{(x,y)} = \frac{a^2y}{b^2x}$$So, the equation of normal to the curve (1) at $(x,y)$ is:
$$Y - y = \left[-\frac{dx}{dy}\right]_{(x,y)}(X - x)$$ $$\text{or,} \quad Y - y = \frac{a^2y}{b^2x}(X - x)$$ $$\text{or,} \quad a^2y(X - x) = b^2x(Y - y) \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(a\sec\theta, b\tan\theta)$.
Solution
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \implies b^2x^2 - a^2y^2 = a^2b^2 \quad \longrightarrow(1)$$
Differentiating (1) w.r.t. $x$, we get:
$$b^2 \cdot 2x - a^2 \cdot 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{b^2x}{a^2y}$$ $$\therefore \left[-\frac{dx}{dy}\right]_{(a\sec\theta, b\tan\theta)} = -\left[\frac{a^2y}{b^2x}\right]_{(a\sec\theta, b\tan\theta)} = -\frac{a^2 \cdot b\tan\theta}{b^2 \cdot a\sec\theta} = -\frac{a}{b}\sin\theta$$So, the equation of normal to the curve (1) at $(a\sec\theta, b\tan\theta)$ is:
$$y - b\tan\theta = -\frac{a}{b}\sin\theta (x - a\sec\theta)$$ $$\text{or,} \quad by - b^2\tan\theta = -ax\sin\theta + a^2\sin\theta \cdot \frac{1}{\cos\theta}$$ $$\text{or,} \quad by - b^2\tan\theta = -ax\sin\theta + a^2\tan\theta$$ $$\text{or,} \quad ax\sin\theta + by = (a^2 + b^2)\tan\theta$$Dividing both sides by $\tan\theta$ gives the standard form:
$$ax\cos\theta + by\cot\theta = a^2 + b^2 \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $x^2 + y^2 - 4x - 6y + 3 = 0$ at $(1,6)$.
Solution
$$x^2 + y^2 - 4x - 6y + 3 = 0 \quad \longrightarrow(1)$$
Differentiating (1) w.r.t. $x$, we get:
$$2x + 2y\frac{dy}{dx} - 4 - 6\frac{dy}{dx} = 0$$ $$\text{or,} \quad (2x - 4) + (2y - 6)\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{2(x-2)}{2(y-3)} = -\frac{x-2}{y-3}$$ $$\therefore \left[-\frac{dx}{dy}\right]_{(1,6)} = \left[\frac{y-3}{x-2}\right]_{(1,6)} = \frac{6-3}{1-2} = -3$$So, the equation of normal to the curve (1) at $(1,6)$ is:
$$y - 6 = \left[-\frac{dx}{dy}\right]_{(1,6)}(x - 1)$$ $$\text{or,} \quad y - 6 = -3(x - 1) \implies y - 6 = -3x + 3$$ $$\therefore 3x + y = 9 \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $x^2 + y^2 = 8$ at the point where $\frac{dy}{dx} = -1$.
Solution
$$x^2 + y^2 = 8 \quad \longrightarrow(1)$$
Differentiating (1) w.r.t. $x$, we get:
$$2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$$According to the question:
$$\frac{dy}{dx} = -1 \implies -\frac{x}{y} = -1 \implies x = y = h \quad (\text{say})$$ $$\text{Since } \frac{dy}{dx} = -1, \text{ the slope of normal } \left[-\frac{dx}{dy}\right] = 1$$So, the equation of normal to the curve (1) at $(h,h)$ is given by:
$$y - h = 1 \cdot (x - h) \implies y - h = x - h$$ $$\therefore x - y = 0 \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $x^3 + y^3 = 3axy$ at $\left(\frac{3a}{2}, \frac{3a}{2}\right)$.
Solution
$$x^3 + y^3 = 3axy \quad \longrightarrow(1)$$
Differentiating (1) w.r.t. $x$, we get:
$$3x^2 + 3y^2\frac{dy}{dx} = 3a \left(1 \cdot y + x \frac{dy}{dx}\right)$$ $$\text{or,} \quad x^2 + y^2\frac{dy}{dx} = ay + ax\frac{dy}{dx} \implies (y^2 - ax)\frac{dy}{dx} = ay - x^2$$ $$\text{or,} \quad \frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax} \implies -\frac{dx}{dy} = \frac{ax - y^2}{ay - x^2}$$ $$\text{At } \left(\frac{3a}{2}, \frac{3a}{2}\right): \quad \left[-\frac{dx}{dy}\right]_{\left(\frac{3a}{2}, \frac{3a}{2}\right)} = \frac{a \cdot \frac{3a}{2} - \left(\frac{3a}{2}\right)^2}{a \cdot \frac{3a}{2} - \left(\frac{3a}{2}\right)^2} = 1$$So, the equation of normal to the curve (1) at $\left(\frac{3a}{2}, \frac{3a}{2}\right)$ is:
$$y - \frac{3a}{2} = 1 \cdot \left(x - \frac{3a}{2}\right) \implies x - y = 0 \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $x = y^2 - 4y$ at points where the curve crosses the y-axis.
Solution
$$x = y^2 - 4y \quad \longrightarrow(1)$$
Let the curve (1) cross the y-axis at $(0,k)$. Setting $x = 0$:
$$0 = k^2 - 4k \implies k(k - 4) = 0 \implies k = 0 \text{ or } k = 4$$So, the points of intersection are $(0,0)$ and $(0,4)$.
Differentiating (1) w.r.t. $x$, we get:
$$1 = 2y\frac{dy}{dx} - 4\frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{2y - 4}$$ $$\therefore -\frac{dx}{dy} = -(2y - 4) = 4 - 2y$$Case 1: At $(0,0)$
$$\left[-\frac{dx}{dy}\right]_{(0,0)} = 4 - 2(0) = 4$$ $$\text{Normal Eq: } y - 0 = 4(x - 0) \implies 4x - y = 0 \quad \mathbf{\color{#059669}{(Ans)}}$$Case 2: At $(0,4)$
$$\left[-\frac{dx}{dy}\right]_{(0,4)} = 4 - 2(4) = -4$$ $$\text{Normal Eq: } y - 4 = -4(x - 0) \implies 4x + y = 4 \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $x = at^2, y = 2at$ at the point $t$.
Solution
$$x = at^2 \implies \frac{dx}{dt} = 2at$$
$$y = 2at \implies \frac{dy}{dt} = 2a$$
$$\therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t} \implies -\frac{dx}{dy} = -t$$
Hence, the equation of normal to the given curve at point $t$ is:
$$y - 2at = -t(x - at^2) \implies y - 2at = -tx + at^3$$ $$\therefore tx + y = 2at + at^3 \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $x = a(2\cos t + \cos 2t), y = a(2\sin t - \sin 2t)$ at $t = \frac{\pi}{2}$.
Solution
$$x = a(2\cos t + \cos 2t) \implies \frac{dx}{dt} = a(-2\sin t - 2\sin 2t) = -2a(\sin t + \sin 2t)$$
$$y = a(2\sin t - \sin 2t) \implies \frac{dy}{dt} = a(2\cos t - 2\cos 2t) = 2a(\cos t - \cos 2t)$$
$$\therefore \frac{dy}{dx} = \frac{2a(\cos t - \cos 2t)}{-2a(\sin t + \sin 2t)} = \frac{\cos 2t - \cos t}{\sin t + \sin 2t}$$
$$\text{So,} \quad -\frac{dx}{dy} = \frac{\sin t + \sin 2t}{\cos t - \cos 2t}$$
$$\therefore \left[-\frac{dx}{dy}\right]_{t=\pi/2} = \frac{\sin(\pi/2) + \sin\pi}{\cos(\pi/2) - \cos\pi} = \frac{1 + 0}{0 - (-1)} = 1$$
Calculating point coordinates at $t = \frac{\pi}{2}$:
$$x = a\left(2\cos\frac{\pi}{2} + \cos\pi\right) = a(0 - 1) = -a$$ $$y = a\left(2\sin\frac{\pi}{2} - \sin\pi\right) = a(2 - 0) = 2a$$Hence, the equation of normal at $t = \frac{\pi}{2}$ is:
$$y - 2a = 1 \cdot (x + a) \implies y - 2a = x + a$$ $$\therefore x - y + 3a = 0 \quad \mathbf{\color{#059669}{(Ans)}}$$
Find the equation of the normal to $x = a\sin^3t, y = b\cos^3t$ at the point $t$.
Solution
$$x = a\sin^3t \implies \frac{dx}{dt} = 3a\sin^2t\cos t$$
$$y = b\cos^3t \implies \frac{dy}{dt} = -3b\cos^2t\sin t$$
$$\therefore \frac{dy}{dx} = \frac{-3b\cos^2t\sin t}{3a\sin^2t\cos t} = -\frac{b\cos t}{a\sin t} \implies -\frac{dx}{dy} = \frac{a\sin t}{b\cos t}$$
Hence, the equation of normal at point $t$ is:
$$y - b\cos^3t = \frac{a\sin t}{b\cos t} (x - a\sin^3t)$$ $$\text{or,} \quad b\cos t (y - b\cos^3t) = a\sin t (x - a\sin^3t)$$ $$\text{or,} \quad by\cos t - b^2\cos^4t = ax\sin t - a^2\sin^4t$$ $$\therefore ax\sin t - by\cos t = a^2\sin^4t - b^2\cos^4t \quad \mathbf{\color{#059669}{(Ans)}}$$
Please do not enter any spam link in the comment box