Pam Stokes

Answered

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.

Answer & Explanation

Chanell Sanborn

Expert

2021-12-22Added 41 answers

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$2a=16\Rightarrow a=8$ ,

Let the length of minor axis is$=2b\text{}with\text{}\left(0b8\right)$

Therefore equation of required ellipse is given by

$\frac{{x}^{2}}{{8}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\left(8>b\right)$

$\Rightarrow \frac{{x}^{2}}{64}+\frac{{y}^{2}}{{b}^{2}}=1$ (i)

Step 3

Now differentiating the equation (i) with respect to x,

$\frac{d}{dx}(\frac{{x}^{2}}{64}+\frac{{y}^{2}}{{b}^{2}})=\frac{d}{dx}\left(1\right)$

$\Rightarrow \frac{d}{dx}\left(\frac{{x}^{2}}{64}\right)+\frac{d}{dx}\left(\frac{{y}^{2}}{{b}^{2}}\right)=0$ (differentiation of constant is zero)

$\Rightarrow \left(\frac{2x}{64}\right)+\frac{d}{dx}\left(\frac{{y}^{2}}{{b}^{2}}\right)=0\text{}(\mathrm{sin}ce\text{}\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$ )

$\Rightarrow \frac{x}{32}+\frac{d}{dy}\left(\frac{{y}^{2}}{{b}^{2}}\right)\frac{dy}{dx}=0$ (by chain rule)

$\Rightarrow \frac{x}{32}+\left(\frac{2y}{{b}^{2}}\right)\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{x\times {b}^{2}}{32\times 2y}$

$\Rightarrow \frac{dy}{dx}=-\frac{x{b}^{2}}{64y}$ (ii)

From equation (i), we have

To determine: The differential equation which will satisfy the above mentioned set up.

Step 2

Explanation:

It is known that "The ellipse with equation

As length of major axis of the given ellipse is 16.

Therefore

Let the length of minor axis is

Therefore equation of required ellipse is given by

Step 3

Now differentiating the equation (i) with respect to x,

From equation (i), we have

Piosellisf

Expert

2021-12-23Added 40 answers

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

$\frac{x}{32}+\frac{2y{y}^{\prime}}{{b}^{2}}=0$

$b}^{2}=\frac{64y{y}^{\prime}}{x$

$\frac{{x}^{2}}{{8}^{2}}+\frac{xy}{64{y}^{\prime}}=1$

$\frac{xy}{{y}^{\prime}}=64-{x}^{2}$

$y}^{\prime}=\frac{xy}{64-{x}^{2}$

Given that center of ellipse is (0, 0) major axis is 16, semi-minor axis is b ,

so equation of ellipse will

we have to find differential equation which will satisfy above mentioned set up.

Step 2

now

differentiating with respect to x

RizerMix

Expert

2021-12-29Added 437 answers

Where

Satisfy

Suppose the solution is

Then, calculating the differential of both sides

So we see

We can integrate these equations partially to find f. let's go for the y integral

where x(x) is an arbitrary function dx only.

Now substitute that into the equation

for

Comparing what with the coefficient of dx we get that

So the solution is F=c or

Most Popular Questions