Question

The equation of the normal to the curve 3x² − y² = 8 which is parallel to x + 3y = 8 is ____________ .

Options

• x + 3y = 8

• x + 3y + 8 = 0

• x + 3y ± 8 = 0

• x + 3y = 0

Answer 1

x + 3y ± 8 = 0

 

The slope of line x + 3y = 8 is $$\frac{-1}{3}$$.

$$\text{ Since the normal is parallel to the given line, the equation of normal will be of the given form.}$$

$$x+3y=k$$

$$3x^2-y^2=8$$

$$\text{ Let}(x_1,y_1)\text{ be the point of intersection of the two curves}.$$

$$\text{ Then },$$

$$x_1+3y_1=k...\left(1\right)$$

$$3{x_1}^2-{y_1}^2=8...\left(2\right)$$

$$\text{ Now, }3x^2-y^2=8$$

$$\text{ On differentiating both sides w.r.t.x, we get }$$

$$6x-2y\frac{dy}{dx}=0$$

$$\Rightarrow \frac{dy}{dx}=\frac{6x}{2y}=\frac{3x}{y}$$

$$\text{ Slope of the tangent }={\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\frac{3x_1}{y_1}$$

$$\text{ Slope of the normal },m=\frac{-1}{\left(\frac{3x}{y}\right)}=\frac{-y_1}{3x_1}$$

$$\text{ Given }:$$

$$\text{ Slope of the normal = Slope of the given line }$$

$$\Rightarrow \frac{-y_1}{3x_1}=\frac{-1}{3}$$

$$\Rightarrow y_1=x_1...\left(3\right)$$

$$\text{ From }(2),\text{ we get }$$

$$3{x_1}^2-{x_1}^2=8$$

$$\Rightarrow 2{x_1}^2=8$$

$$\Rightarrow {x_1}^2=4$$

$$\Rightarrow x_1=\pm 2$$

$$\text{ Case }1:$$

$$\text{ When }{x}_1=2$$

$$\text{ From (3), we get }$$

$$y_1=x_1=2$$

$$\therefore (x_1,y)=(2,2)$$

$$\text{ From (1), we get }$$

$$2+3\left(2\right)=k$$

$$\Rightarrow 2+6=k$$

$$\Rightarrow k=8$$

$$\therefore \text{ Equation of the normal from}(1)$$

$$\Rightarrow x+3y=8$$

$$\Rightarrow x+3y-8=0$$

$$\text{ Case }2:$$

$$\text{ When }{x}_1=-2$$

$$\text{ From (3), we get }$$

$$y_1=x_1=-2,$$

$$\therefore (x_1,y)=(-2,-2)$$

$$\text{ From (1), we get }$$

$$-2+3(-2)=k$$

$$\Rightarrow -2-6=k$$

$$\Rightarrow k=-8$$

$$\therefore \text{ Equation of the normal from }(1)$$

$$\Rightarrow x+3y=-8$$

$$\Rightarrow x+3y+8=0$$

$$\text{ From both the cases, we get the equation of the normal as }:$$

$$x+3y\pm 8=0$$