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प्रश्न
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
उत्तर
x + 3y ± 8 = 0
The slope of line x + 3y = 8 is \[\frac{- 1}{3}\].
\[\text { Since the normal is parallel to the given line, the equation of normal will be of the given form.}\]
\[x + 3y = k \]
\[3 x^2 - y^2 = 8 \]
\[\text { Let} \left( x_1 , y_1 \right) \text { be the point of intersection of the two curves} . \]
\[\text { Then }, \]
\[ x_1 + 3 y_1 = k . . . \left( 1 \right)\]
\[3 {x_1}^2 - {y_1}^2 = 8 . . . \left( 2 \right)\]
\[\text { Now, } 3 x^2 - y^2 = 8 \]
\[\text { On differentiating both sides w.r.t.x, we get }\]
\[6x - 2y\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{6x}{2y} = \frac{3x}{y}\]
\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{3 x_1}{y_1}\]
\[\text { Slope of the normal }, m=\frac{- 1}{\left( \frac{3x}{y} \right)}=\frac{- y_1}{3 x_1}\]
\[\text { Given }:\]
\[\text { Slope of the normal = Slope of the given line }\]
\[ \Rightarrow \frac{- y_1}{3 x_1} = \frac{- 1}{3}\]
\[ \Rightarrow y_1 = x_1 . . . \left( 3 \right)\]
\[\text { From } (2), \text { we get }\]
\[3 {x_1}^2 - {x_1}^2 = 8\]
\[ \Rightarrow 2 {x_1}^2 = 8\]
\[ \Rightarrow {x_1}^2 = 4\]
\[ \Rightarrow x_1 = \pm 2\]
\[\text { Case } 1:\]
\[\text { When } x_1 =2\]
\[\text { From (3), we get }\]
\[ y_1 = x_1 = 2\]
\[ \therefore \left( x_1 , y \right) = \left( 2, 2 \right)\]
\[\text { From (1), we get }\]
\[2 + 3\left( 2 \right) = k\]
\[ \Rightarrow 2 + 6 = k\]
\[ \Rightarrow k = 8\]
\[ \therefore \text { Equation of the normal from} (1)\]
\[ \Rightarrow x + 3y = 8\]
\[ \Rightarrow x + 3y - 8 = 0\]
\[\text { Case } 2:\]
\[\text { When } x_1 =-2\]
\[\text { From (3), we get }\]
\[ y_1 = x_1 = - 2, \]
\[ \therefore \left( x_1 , y \right) = \left( - 2, - 2 \right)\]
\[\text { From (1), we get }\]
\[ - 2 + 3\left( - 2 \right) = k\]
\[ \Rightarrow - 2 - 6 = k\]
\[ \Rightarrow k = - 8\]
\[ \therefore \text { Equation of the normal from } (1)\]
\[ \Rightarrow x + 3y = - 8\]
\[ \Rightarrow x + 3y + 8 = 0\]
\[\text { From both the cases, we get the equation of the normal as }:\]
\[x + 3y \pm 8 = 0\]
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