 Pam Stokes

2021-12-21

A certain ellipse is centered at (0, 0) from which the major axis always equal to 16 and the minor axis is free to change as long as it obeys $a>b$ for ellipses. The directrices of this ellipse is perpendicular to the x - axis. Determine the differential equation which will satisfy the abovementioned set up. Chanell Sanborn

Expert

Step 1
To determine: The differential equation which will satisfy the above mentioned set up.
Step 2
Explanation:
It is known that "The ellipse with equation $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\left(a>b\right)$ has center at (0, 0), major axis 2a, minor axis 2b, and directrices perpendicular to the x-axis."
As length of major axis of the given ellipse is 16.
Therefore $2a=16⇒a=8$,
Let the length of minor axis is
Therefore equation of required ellipse is given by
$\frac{{x}^{2}}{{8}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\left(8>b\right)$
$⇒\frac{{x}^{2}}{64}+\frac{{y}^{2}}{{b}^{2}}=1$ (i)
Step 3
Now differentiating the equation (i) with respect to x,
$\frac{d}{dx}\left(\frac{{x}^{2}}{64}+\frac{{y}^{2}}{{b}^{2}}\right)=\frac{d}{dx}\left(1\right)$
$⇒\frac{d}{dx}\left(\frac{{x}^{2}}{64}\right)+\frac{d}{dx}\left(\frac{{y}^{2}}{{b}^{2}}\right)=0$ (differentiation of constant is zero)
)
$⇒\frac{x}{32}+\frac{d}{dy}\left(\frac{{y}^{2}}{{b}^{2}}\right)\frac{dy}{dx}=0$ (by chain rule)
$⇒\frac{x}{32}+\left(\frac{2y}{{b}^{2}}\right)\frac{dy}{dx}=0$
$⇒\frac{dy}{dx}=-\frac{x×{b}^{2}}{32×2y}$
$⇒\frac{dy}{dx}=-\frac{x{b}^{2}}{64y}$ (ii)
From equation (i), we have Piosellisf

Expert

Step 1
Given that center of ellipse is (0, 0) major axis is 16, semi-minor axis is b , $a=8,a>b$ directrices are perpendicular to the x-axis.
so equation of ellipse will $\frac{{x}^{2}}{{8}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$
we have to find differential equation which will satisfy above mentioned set up.
Step 2
now $\frac{{x}^{2}}{{8}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$
differentiating with respect to x
$\frac{x}{32}+\frac{2y{y}^{\prime }}{{b}^{2}}=0$
${b}^{2}=\frac{64y{y}^{\prime }}{x}$
$\frac{{x}^{2}}{{8}^{2}}+\frac{xy}{64{y}^{\prime }}=1$
$\frac{xy}{{y}^{\prime }}=64-{x}^{2}$
${y}^{\prime }=\frac{xy}{64-{x}^{2}}$ RizerMix

Expert

Where
Satisfy $\frac{df}{dy}=\frac{dg}{dx}$
Suppose the solution is $F\left(x,y\right)=c$
Then, calculating the differential of both sides
$\frac{dF}{dx}dx+\frac{dF}{dy}dy=0$
So we see $f=\frac{dF}{dx}$
$g=\frac{dF}{dy}$
We can integrate these equations partially to find f. let's go for the y integral
$\frac{dF}{dy}={y}^{2}+{x}^{2}/\left(1-xy{\right)}^{2}$
$F={y}^{3}/3-{x}^{2}/\left(1-xy\right)\left(-1/x\right)+1x\left(x\right)$
${y}^{3}/3+x/\left(1-xy\right)+x\left(x\right)$
where x(x) is an arbitrary function dx only.
Now substitute that into the equation
for $dF/dx$ to determine x(x)
$\frac{dF}{dx}=1/\left(1-xy{\right)}^{2}+dx/dx$
Comparing what with the coefficient of dx we get that is just a constant which we can take to be 0.
So the solution is F=c or
${y}^{3}/3+x/\left(1-xy\right)=c$
$⇒{y}^{3}+3x/\left(1-xy\right)=3c$
$⇒\left(1-xy\right){y}^{3}+3x=3c\left(1-xy\right)$
$⇒{y}^{3}-x{y}^{4}+3x-3c×4=-3c$
$⇒x{y}^{4}-{y}^{3}+3cxy-3x=3c$

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