Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the equation of the normal lines to the curve 3x^2 – y^2 = 8 which are parallel to the line x + 3y = 4.

Find the equation of the normal lines to the curve 3x^2 - y^2 = 8  which are parallel to the line x + 3y = 4.

Answers (1)

Given: equation of the curve  3x^2 - y^2 = 8 equation of line  x + 3y = 4

To find: the equation of the normal lines to the given curve which are parallel to the given line

Explanation:

Now given equation of curve as  3x^2 - y^2 = 8

Differentiating the equation with respect to X

\\\frac{\mathrm{d}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(8)}{\mathrm{dx}}$ \\$\Rightarrow \frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{d} \mathrm{x}}=0$ \\$\Rightarrow 3 \times 2 \mathrm{x}-2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0$ \\$\Rightarrow 6 \mathrm{x}=2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6 \mathrm{x}}{2 \mathrm{y}}=\frac{3 \mathrm{x}}{\mathrm{y}}=\mathrm{m}_{1} \ldots \ldots(\mathrm{i})

Assume the slope of the normal curve to be m2 is given by

\Rightarrow \mathrm{m}_{2}=\frac{-1}{\mathrm{~m}_{1}}
Substituting value from equation (i), we get
\\\Rightarrow \mathrm{m}_{2}=\frac{-1}{\frac{3 \mathrm{x}}{\mathrm{y}}} \\\Rightarrow \mathrm{m}_{2}=\frac{-\mathrm{y}}{3 \mathrm{x}}
The known equation of the line is
\\$x+3 y=4 \\ $\Rightarrow 3 y=4-x

\\\Rightarrow \mathrm{y}=\frac{4}{3}-\frac{\mathrm{x}}{3}$\\ $\Rightarrow \mathrm{y}=-\frac{\mathrm{x}}{3}+\frac{4}{3}
\therefore  the slope of this line will be
\mathrm{m}_{3}=-\frac{1}{3} \ldots \ldots (iii)
since, slope of normal to the curve should be equal to the slope of the line which is parallel to the curve,
\therefore \mathrm{m}_{2}=\mathrm{m}_{3}
Substituting values from equation (ii) and (iii), we get

\\ \Rightarrow-\frac{y}{3 x}=-\frac{1}{3}$ \\$\Rightarrow 3 y=3 x$ \\$\Rightarrow y=x_{\cdots \cdots}(i v)
After putting y=x in the equation of the curve, we get
\\3 x^{2}-y^{2}=8$ \\$\Rightarrow 3 x^{2}-x^{2}=8$ \\$\Rightarrow 2 x^{2}=8$ \\$\Rightarrow x^{2}=4$ \\$\Rightarrow x=\pm 2
But from equation (iv)
y=x
\Rightarrow y=\pm 2

Therefore, the points at which normal to the given curve is parallel to the given line are (2, 2) and (-2, -2)

Thus, the equations of the normal can be calculated by

\begin{aligned} &y-2=m_{2}(x-2) \text { and } y+2=m_{2}(x+2)\\ &\Rightarrow \mathrm{y}-2=-\frac{1}{3}(\mathrm{x}-2) \quad \text{And }\mathrm{y}+2=-\frac{1}{3}(\mathrm{x}+2)\\ &3 y-6=-x+2 \text { and } 3 y+6=-x-2\\ &3 y+x=8 \text { and } 3 y+x=-8\\ &\text { Hence the equation of the normal lines to the curve } 3 x^{2}-y^{2}=8 \text { which are parallel to the line }\\ &x+3 y=8 \text { and } 3 y+y=- 8 \end{aligned}

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question