A certain ellipse is centered at (0, 0) from which the major axis always equal...

Pam Stokes

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

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 x2a2+y2b2=1(a>b) 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=16a=8,
Let the length of minor axis is =2b with (0<b<8)
Therefore equation of required ellipse is given by
x282+y2b2=1(8>b)
x264+y2b2=1 (i)
Step 3
Now differentiating the equation (i) with respect to x,
ddx(x264+y2b2)=ddx(1)
ddx(x264)+ddx(y2b2)=0 (differentiation of constant is zero)
(2x64)+ddx(y2b2)=0 (since ddx(xn)=nxn1)
x32+ddy(y2b2)dydx=0 (by chain rule)
x32+(2yb2)dydx=0
dydx=x×b232×2y
dydx=xb264y (ii)
From equation (i), we have
Piosellisf

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 x282+y2b2=1
we have to find differential equation which will satisfy above mentioned set up.
Step 2
now x282+y2b2=1
differentiating with respect to x
x32+2yyb2=0
b2=64yyx
x282+xy64y=1
xyy=64x2
y=xy64x2
RizerMix

RizerMix

Expert

2021-12-29Added 437 answers

f dx1gdy=0
Where f=(1xy)(2) and 9=(y2+x2(1xy)(2))
Satisfy dfdy=dgdx
Suppose the solution is F(x,y)=c
Then, calculating the differential of both sides
dFdxdx+dFdydy=0
So we see f=dFdx
g=dFdy
We can integrate these equations partially to find f. let's go for the y integral
dFdy=y2+x2/(1xy)2
F=y3/3x2/(1xy)(1/x)+1x(x)
y3/3+x/(1xy)+x(x)
where x(x) is an arbitrary function dx only.
Now substitute that into the equation
for dF/dx to determine x(x)
dFdx=1/(1xy)2+dx/dx
Comparing what with the coefficient of dx we get that dx/dx=0 so x(x) is just a constant which we can take to be 0.
So the solution is F=c or
y3/3+x/(1xy)=c
y3+3x/(1xy)=3c
(1xy)y3+3x=3c(1xy)
y3xy4+3x3c×4=3c
xy4y3+3cxy3x=3c

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?