Cfrgtkky

The question is as follows:

$$y=x^2/(x+1)$$

The normal to this curve at $x=1$ meets the $x$-axis at point $M$.

The tangent to the curve at $x=-2$ meets the $y$-axis at point $N$.

Find the area of triangle $MNO$, where $O$ is the origin.

PS- this is not a school h.w so don't worry. And I did try....for a good 100 min...not even joking

EDIT - The drawing is NOT EXACT, it is just to give an idea.

$y(x) = dfrac{x^2}{x + 1} = (x + 1)^{-1}x^2; tag 1$

$y'(x) = -(x + 1)^{-2}x^2 + 2x(x + 1)^{-1} = -(x + 1)^{-2}x^2 + 2x(x + 1)(x + 1)^{-2}$
$= -(x + 1)^{-2}x^2 + (2x^2 + 2x)(x + 1)^{-2} = (x^2 + 2x)(x + 1)^{-2} = dfrac{x^2 + 2x}{(x + 1)^2}; tag 2$

$y(1) = dfrac{1}{2}; ; y'(1) = dfrac{3}{4}; tag 3$

$y(-2) = -4; ; y'(-2) = 0; tag 4$

the slope of the normal line through $(1, y(1)) = (1, 1/2)$ is then

$m = -dfrac{1}{y'(1)} = -dfrac{4}{3}; tag 5$

the equation of the normal line through $(1, 1/2)$ is thus

$y - dfrac{1}{2} = -dfrac{4}{3}(x - 1), tag 6$

which meets the $x$-axis where $y = 0$:

$-dfrac{1}{2} = -dfrac{4}{3}(x - 1) Longrightarrow x = dfrac{3}{8} + 1 = dfrac{11}{8}; tag 7$

thus,

$M = left (dfrac{11}{8}, 0 right); tag 8$

likewise, the tangent line through $(-2, y(-2)) = (-2, -4)$ is

$y + 4 = 0(x + 2) = 0, tag 9$

which intersects the $y$-axis where $x = 0$, with $y$-coordinate given by

$y + 4 = 0 Longrightarrow y = -4, tag{10}$

and so

$N = left (0, -4 right ); tag{11}$

the area $A$ of $triangle MNO$ is thus

$A = dfrac{1}{2}ON cdot OM = dfrac{1}{2} 4 cdot dfrac{11}{8} = dfrac{11}{4}. tag{12}$

Given $y=x^2/(x+1)$, and $f'(frac{u}v) = (u'v-v'u)/v^2 $
$$y' = frac{2x(x+1)-(x^2)}{(x^2+1)^2} = frac{x^2+2x}{(x+1)^2}$$

The normal line's slope is $-frac1{y'}=-frac{(x+1)^2}{x^2+2x}$. For the function, $y(1) = frac12$ and $y(-2)=-4$

The slope of the normal through $(1,frac12)$ is $-frac43$, so the line is $y_1=-frac43(x-1)+frac12$. This has an x-intercept of $x-1=frac34 cdot frac12 to x=frac{11}8$.

The slope of the tangent through $(-2,-4)$ is $0$, so the y-intercept will be $-4$.

The triangle now has a base $|M-0|$ of $frac{11}8$ and a height $|N-0|$ of $4$.

Area = $$frac12 bh=frac{1*11*4}{2*8*1}=frac{11}4$$