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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.

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

Answers (1)

Answer:

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

Hint:

          Use differentiation

Given:

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

Solution:

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{dy}{dx}}

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

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

3y=8-x

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,

3x^{2}-y^{2}=8

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

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

 

12-8=y^{2}

y^{2}=4

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

 

                

Posted by

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