Question

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

विकल्प

• 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 \[\frac{- 1}{3}\].

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

\[x + 3y = k \]

\[3 x^2 - y^2 = 8 \]

\[\text { Let} \left( x_1 , y_1 \right) \text { be the point of intersection of the two curves} . \]

\[\text { Then }, \]

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

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

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

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

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

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

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

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

\[\text { Given }:\]

\[\text { Slope of the normal = Slope of the given line }\]

\[ \Rightarrow \frac{- y_1}{3 x_1} = \frac{- 1}{3}\]

\[ \Rightarrow y_1 = x_1 . . . \left( 3 \right)\]

\[\text { From } (2), \text { we get }\]

\[3 {x_1}^2 - {x_1}^2 = 8\]

\[ \Rightarrow 2 {x_1}^2 = 8\]

\[ \Rightarrow {x_1}^2 = 4\]

\[ \Rightarrow x_1 = \pm 2\]

\[\text { Case } 1:\]

\[\text { When } x_1 =2\]

\[\text { From (3), we get }\]

\[ y_1 = x_1 = 2\]

\[ \therefore \left( x_1 , y \right) = \left( 2, 2 \right)\]

\[\text { From (1), we get }\]

\[2 + 3\left( 2 \right) = k\]

\[ \Rightarrow 2 + 6 = k\]

\[ \Rightarrow k = 8\]

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

\[ \Rightarrow x + 3y = 8\]

\[ \Rightarrow x + 3y - 8 = 0\]

\[\text { Case } 2:\]

\[\text { When } x_1 =-2\]

\[\text { From (3), we get }\]

\[ y_1 = x_1 = - 2, \]

\[ \therefore \left( x_1 , y \right) = \left( - 2, - 2 \right)\]

\[\text { From (1), we get }\]

\[ - 2 + 3\left( - 2 \right) = k\]

\[ \Rightarrow - 2 - 6 = k\]

\[ \Rightarrow k = - 8\]

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

\[ \Rightarrow x + 3y = - 8\]

\[ \Rightarrow x + 3y + 8 = 0\]

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

\[x + 3y \pm 8 = 0\]