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Need Solution for R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question Question 10 Maths Textbook Solution.

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                \text { (c) } x+3 y \pm 8=0


          Use differentiation


            3x^{2}-y^{2}=8 Is parallel to x+3y=8


Given the equation of the line 3x^{2}-y^{2}=8 now differentiating both sides with respect to x, we get

\frac{d\left ( 3x^{2}-y^{2} \right )}{dx}=\frac{d\left ( 8 \right )}{dx}

Now applying the sum rule of differentiation an differentiation of constant =0,so we get

\begin{aligned} &\frac{d\left(3 x^{2}\right)}{d x}-\frac{d\left(y^{2}\right)}{d x}=0 \\ &3 \frac{d\left(x^{2}\right)}{d x}-\frac{d\left(y^{2}\right)}{d x}=0 \\ &3 \times 2\left(x^{2-1}\right) \times \frac{d(x)}{d x}-2\left(y^{2-1}\right) \times \frac{d(y)}{d x}=0 \\ &6 x-2 y \times \frac{d y}{d x}=0 \\ &6 x=2 y \times \frac{d y}{d x} \\ &\frac{d y}{d x}=\frac{6 x}{2 y}=\frac{3 x}{y} \end{aligned}

So, this is the slope of the given curve. We know the slope of the normal to the curve is


=-\frac{1}{\frac{3x}{y}}=\left ( -\frac{y}{3x} \right )                                                        (1)

Now the given equation of the line x+3y=8


Differentiating w.r.t x we get

\begin{aligned} &\frac{d(3 y)}{d x}=\frac{d(8-x)}{d x} \\ &3 \frac{d y}{d x}=-1 \\ &\frac{d y}{d x}=-\frac{1}{3} \end{aligned}

So, the slope of the line is -\frac{1}{3}

Now, as the normal to the curve is parallel to this line, hence the slope of the line should be equal to the slope of the normal to the given curve,

\begin{aligned} &\therefore\left(-\frac{y}{3 x}\right)=-\frac{1}{3} \\ &3 y=3 x \\ &y=x \end{aligned}

On substituting this value of the given equation of the curve, we get

\begin{aligned} &3 x^{2}-y^{2}=8 \\ &3 x^{2}-(x)^{2}=8 \\ &2 x^{2}=8 \\ &x^{2}=4 \\ &x=\pm 2 \end{aligned}

When x=2 the equation of the curve becomes,


3\left ( 2 \right )^{2}-y^{2}=8

3\left ( 4 \right )-y^{2}=8




y=\pm 2

When x=-2, the equation of the curve becomes,

\begin{aligned} &3 x^{2}-y^{2}=8 \\ &3(-2)^{2}-y^{2}=8 \\ &3(4)-y^{2}=8 \\ &12-8=y^{2} \\ &y^{2}=4 \\ &y=\pm 2 \end{aligned}

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

And required equation of the normal to the curve at \left ( \pm 2,\pm 2 \right ) is

\begin{aligned} &y-(\pm 2)=-\frac{1}{3}[x-(\pm 2)] \\ &3(y-(\pm 2))=-(x-(\pm 2)) \\ &3 y-(\pm 6)=-x+(\pm 2) \\ &x+3 y-(\pm 6)-(\pm 2)=0 \\ &x+3 y+(\pm 8)=0 \end{aligned}

Hence the equation of normal to the curve

3x^{2}-y^{2}=8 Which is parallel to the line  x+3y=8 is  x+3y+\left ( \pm 8 \right )=0



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