William Burnett

2021-12-29

Consider the curve created by $2{x}^{2}+3{y}^{2}–4xy=36$
(a) Show that $\frac{dy}{dx}=2y-2x3y-2x$
(b) Calculate the slope of the line perpendicular to the curve at each location where x is. $x=6$.
(c) A vertical tangent line to the curve exists at a positive value of x.

usumbiix

So,
$2{x}^{2}+3{y}^{2}-4xy=36$

let $u=3{y}^{2}$

Corgnatiui

Part B : slope of the line tangent to the curve at each point on the curve where $x=6$.
The slope of the curve will be $\frac{dy}{dx}=\frac{2y-2x}{3y-2x}$
we need to find slope of line tangent to curve at $x=6$
first we need find y value at $x=6$. For that substitute $x=6$ in equation of curve.
$2{x}^{2}+3{y}^{2}-4xy=36$
$2\left({6}^{2}\right)+3{y}^{2}-4×6×y=36$
$72+3{y}^{2}-24y-36=0$
$3{y}^{2}-24y+36=0$
divide by 3 we get,
${y}^{2}-8y+12=0$
$\left(y-6\right)\left(y-2\right)=0$
therefore $y=6$ or $y=2$
therefore the point off tangency are
$\left(x,y\right)=\left(6,6\right)$ and $\left(x,y\right)=\left(6,2\right)$
equation of slope $=\frac{dy}{dx}=\frac{2y-2x}{3y-2x}=0$
at $\left(x,y\right)=\left(6,6\right)$
slope ${m}_{1}=\frac{2×6-2×6}{3×6-2×6}=0$
${m}_{1}=0$
$\left(x,y\right)=\left(6,2\right)$
slope ${m}_{2}=\frac{2×2-2×6}{3×2-2×6}=\frac{4-12}{6-12}=\frac{8}{6}=\frac{4}{3}$
${m}_{2}=\frac{4}{3}$

karton

Part C : value of x at which the curve has a vertical tangent line
The vertical tangent to a curve occurs at a point where slope is undefined or infinite.
This is nothing but the points where derivative of function is not defined.
we know the derivative of function is,
$\frac{dy}{dx}=\frac{2y-2x}{3y-2x}$
$\frac{dy}{dx}$ is undefined at points where the denominator becomes zero.
that is 3y − 2x = 0
3y = 2x
$y=\frac{2}{3}x$
substitute this in equation of curve,
$\begin{array}{}2{x}^{2}+3{y}^{2}-4xy=36\\ 2{x}^{2}+3\left(\frac{2}{3}x{\right)}^{2}-4x\left(\frac{2}{3}x\right)=36\\ 2{x}^{2}+\frac{4}{3}{x}^{2}-\frac{8}{3}{x}^{2}=36\\ 2{x}^{2}-\frac{4}{3}{x}^{2}=36\\ \text{multiply throughiut by 3,}\\ 6{x}^{2}-4{x}^{2}=108\\ 2{x}^{2}=108\\ {x}^{2}=\frac{108}{2}=54\\ x=±\sqrt{54}=±7.348\end{array}$
therefore vertical tangent to the curve occurs at points x = +7.348 and x = - 7.348

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