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What is the equation of the normal line of $f (x) = \frac{x^{2}}{1 + 4 x}$ $f(x)=x^2/(1+4x)$ at $x = - 1$ $x=-1$ ?

1 Answer

Shwetank Mauria

Mar 26, 2016

Equation of normal at $x = - 1$ $x=-1$ is $27 x + 6 y + 29 = 0$ $27x+6y+29=0$

Explanation:

As $f (x) = \frac{x^{2}}{1 + 4 x}$ $f(x)=x^2/(1+4x)$ , at $x = - 1$ $x=-1$ , we have $f (- 1) = \frac{{(- 1)}^{2}}{1 + 4 (- 1)} = \frac{1}{- 3} = - \frac{1}{3}$ $f(-1)=(-1)^2/(1+4(-1))=1/-3=-1/3$

And hence normal passes through $(- 1, - \frac{1}{3})$ $(-1,-1/3)$

Now as $f (x) = \frac{x^{2}}{1 + 4 x}$ $f(x)=x^2/(1+4x)$ ,

$\frac{d f}{d x} = \frac{(1 + 4 x) \times 2 x - 4 \times x^{2}}{{(1 + 4 x)}^{2}} = \frac{(2 x + 8 x^{2}) - 4 x^{2}}{{(1 + 4 x)}^{2}}$ $(df)/(dx)=((1+4x)xx2x-4xxx^2)/(1+4x)^2=((2x+8x^2)-4x^2)/(1+4x)^2$

$\frac{d f}{d x} = \frac{2 x (1 + 2 x)}{{(1 + 4 x)}^{2}}$ $(df)/(dx)=(2x(1+2x))/(1+4x)^2$

and hence slope of curve i.e. tangent at $x = - 1$ $x=-1$ is

$\frac{2 (- 1) (1 + 2 (- 1))}{{(1 + 4 (- 1))}^{2}}$ $(2(-1)(1+2(-1)))/(1+4(-1))^2$ or $\frac{(- 2) \times (- 1)}{{(- 3)}^{2}} = \frac{2}{9}$ $((-2)xx(-1))/(-3)^2=2/9$

and slope of normal would be $- \frac{1}{\frac{2}{9}} = - \frac{9}{2}$ $-1/(2/9)=-9/2$

Hence, equation of normal at $x = - 1$ $x=-1$ is given by $(y + \frac{1}{3}) = - \frac{9}{2} \times (x + 1)$ $(y+1/3)=-9/2xx(x+1)$

or $6 (y + \frac{1}{3}) = - 9 \times \frac{6}{2} \times (x + 1)$ $6(y+1/3)=-9xx6/2xx(x+1)$ or $6 y + 2 = - 27 x - 27$ $6y+2=-27x-27$ or

$27 x + 6 y + 29 = 0$ $27x+6y+29=0$

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