Question

Find the equations of the tangents to the curve

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

which are parallel to the

x

-axis.

Accepted Answer

Given equation of the curve is,

$x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ $. . . . . . . . . . . . . (1)$

On differentiating both sides w.r.to $x$ , we get

$\frac{d}{d x} x^{2} + \frac{d}{d x} y^{2} - \frac{d}{d x} 2 x - \frac{d}{d x} 4 y = - \frac{d}{d x} 1$

$2 x + 2 y \frac{d y}{d x} - 2 - 4 \frac{d y}{d x} = 0$

$2 y \frac{d y}{d x} - 4 \frac{d y}{d x} = 2 - 2 x$

$(2 y - 4) \frac{d y}{d x} = 2 - 2 x$

$\frac{d y}{d x} = \frac{2 - 2 x}{2 y - 4}$

$\frac{d y}{d x} = \frac{1 - x}{y - 2}$

$∴$ Slope of tangent at $(x, y) = \frac{d y}{d x} = \frac{1 - x}{y - 2}$

Given, tangent is parallel to $x -$ axis

$\Rightarrow \frac{d y}{d x} = 0$

$\Rightarrow \frac{1 - x}{y - 2} = 0$

$\Rightarrow 1 - x = 0$

$∴ x = 1$

To get value of $y$ , putting value of $x$ in equation $(1)$

$\Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0$

$\Rightarrow {(1)}^{2} + y^{2} - 2 (1) - 4 y + 1 = 0$

$\Rightarrow 1 + y^{2} - 2 - 4 y + 1 = 0$

$\Rightarrow y^{2} - 4 y = 0$

$\Rightarrow y (y - 4) = 0$

$\Rightarrow y - 4 = 0$

$\Rightarrow y = 4, 0$

Therefore, the coordinates of the points are $(1, 0)$ and $(1, 4)$ .

Since the tangents are parallel to $X$ -axis, their equations are of the form $y = k$

If it passes through the point $(1, 0), k = 0$ , and if it passes through the point $(1, 4), k = 4$

Hence, the equations of the tangents are $y = 0$ and $y = 4$ .

Maharashtra Board Solutions for Chapter: Applications of Derivatives, Exercise 1: EXERCISE 2.1

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Applications of Derivatives, Exercise 1: EXERCISE 2.1

Questions from Maharashtra Board Solutions for Chapter: Applications of Derivatives, Exercise 1: EXERCISE 2.1 with Hints & Solutions

Learn and Prepare for any exam you want !