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  • The equation of normal to the curve 3x^2 - y^2 = 8 which is parallel to the line x + 3y = 8 is A. 3x - y = 8

The equation of normal to the curve 3x\textsuperscript{2} - y^2 = 8 which is parallel to the line x + 3y = 8 is
\\A. 3x - y = 8\\ B. 3x + y + 8 = 0\\ C. x + 3y \pm 8 = 0 \\D. x + 3y = 0

Answers (1)

Given the equation of the line is  3x\textsuperscript{2} - y^2 = 8

Differentiate both sides with x and get

\frac{\mathrm{d}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(8)}{\mathrm{dx}}$
Apply sum rule and 0 is the differentiation of constant, so
\Rightarrow \frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}=0$
Take the constants out and get
\Rightarrow 3 \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}=0$
Apply power rule and get
\\\Rightarrow 3 \times 2\left(\mathrm{x}^{2-1}\right) \times \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}-2\left(\mathrm{y}^{2-1}\right) \times \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dx}}=0$ \\$\Rightarrow 6 \mathrm{x}-2 \mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{dx}}=0$ \\$\Rightarrow 6 \mathrm{x}=2 \mathrm{y} \times \frac{\mathrm{dy}}{\mathrm{dx}} \\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6 \mathrm{x}}{2 \mathrm{y}}=\frac{3 \mathrm{x}}{\mathrm{y}}$

Hence, the slope of the given curve is provided.

Also, the slope of the normal to the curve is

=-\frac{1}{\frac{d y}{d x}}$ \\$\Rightarrow=-\frac{1}{\frac{3 \mathrm{x}}{\mathrm{y}}}=\left(-\frac{\mathrm{y}}{3 \mathrm{x}}\right) \ldots \ldots$ (i)
Now, $x+3 y=8$
\Rightarrow 3 y=8-x$
After differentiating with respect to x
\\\Rightarrow \frac{\mathrm{d}(3 \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}(8-\mathrm{x})}{\mathrm{dx}}$ \\$\Rightarrow 3 \frac{\mathrm{dy}}{\mathrm{dx}}=-1$

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{3}$
Therefore, the slope is -\frac{1}{3}$
Now, because the normal to the curve is parallel to this line, that means the slope of the line must be equal to slope of the normal to the given curve,
\therefore\left(-\frac{y}{3 x}\right)=-\frac{1}{3}$

\\ { \Rightarrow 3y=3x}\\ { \Rightarrow y=x}\\

Substitute the value of the given equation

\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3x\textsuperscript{2} - (x)\textsuperscript{2} = 8}\\ { \Rightarrow 2x\textsuperscript{2} = 8}\\ { \Rightarrow x\textsuperscript{2} = 4}\\ { \Rightarrow x= \pm 2}\\

When x=2, the equation is

\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(2)\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(4) - y\textsuperscript{2} = 8}\\ { \Rightarrow 12-8= y\textsuperscript{2}}\\ { \Rightarrow y\textsuperscript{2} = 4}\\ { \Rightarrow y= \pm 2}\\

When x=-2, the equation is

\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(-2)\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(4) - y\textsuperscript{2} = 8}\\ { \Rightarrow 12-8= y\textsuperscript{2}}\\ { \Rightarrow y\textsuperscript{2} = 4}\\ { \Rightarrow y= \pm 2}\\

So, the points are ( \pm 2, \pm 2) at which normal is parallel to the given line.

And required equation at ( \pm 2, \pm 2) is

\\ y-(\pm 2)=-\frac{1}{3}[x-(\pm 2)] \\ \Rightarrow 3(y-(\pm 2))=-(x-(\pm 2)) \\ \Rightarrow 3 y-(\pm 6)=-x+(\pm 2) \\ \Rightarrow x+3 y-(\pm 6)-(\pm 2)=0 \\ \Rightarrow x+3 y+(\pm \quad 8)=0

Hence the equation of normal to the curve is x+3 y+(\pm \quad 8)=0

So the correct answer is option C

Posted by

infoexpert22

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