Obtain the Differential equations: parabolas with vertex and focus on the x-axis.

ka1leE

ka1leE

Answered question

2020-11-10

Obtain the Differential equations: parabolas with vertex and focus on the x-axis.

Answer & Explanation

stuth1

stuth1

Skilled2020-11-11Added 97 answers

Let M denote the family of parabolas whose vertex and focus both on the x-axis and let (a,0) be the focus of a member of the given family, where a is an arbitrary constant. Therefore, equation of family M is y2=4ax...(1) Differentiating both sides of equation with respect to x, we get
2y(dydx)=4ax
Substituting the value of 4a from equation in (1) y2=(2y(dydx))x
y2=2xy(dydx)
y22xy(dydx)=
which is the Differential equations of the family of parabolas.
nick1337

nick1337

Expert2023-06-14Added 777 answers

To obtain the differential equations for parabolas with the vertex and focus on the x-axis, we can start with the standard equation of a parabola with the vertex at the origin. The equation is given by:
y2=4px where p is the distance from the vertex to the focus.
Now, let's consider a parabola with the vertex at (h,0) and the focus at (p+h,0) on the x-axis. To obtain the equation for this parabola, we can make a translation by substituting x with xh. This gives us:
y2=4p(xh)
Expanding the right side, we have:
y2=4px4ph
Simplifying further, we obtain:
y24px+4ph=0
Finally, we can write the differential equation for this parabola by differentiating both sides with respect to x:
dydx(y2)dydx(4px)+dydx(4ph)=0
Using the chain rule and simplifying, we get:
2y·dydx4p+0=0
Thus, the differential equation for parabolas with the vertex and focus on the x-axis is:
2y·dydx4p=0 where p represents the distance from the vertex to the focus.
Don Sumner

Don Sumner

Skilled2023-06-14Added 184 answers

Answer:
dydx=2ax+b
Explanation:
Given:
y=ax2+bx+c where a, b, and c are constants.
For a parabola with its vertex at the origin (0,0) and its focus at (p, 0), we can use the relationship between the vertex and focus to derive the differential equation.
The distance between the vertex and the focus is given by the equation:
p=14a
Substituting this value into the general equation, we have:
y=ax2+bx+14a
Taking the derivative of both sides with respect to x, we get:
dydx=2ax+b
Therefore, the differential equation for a parabola with its vertex and focus on the x-axis is:
dydx=2ax+b
Vasquez

Vasquez

Expert2023-06-14Added 669 answers

Since the vertex and focus of the parabola are both on the x-axis, we can assume that the vertex is at the origin (0, 0) and the focus is at a point (f, 0).
Using the definition of the parabola, we know that the distance from any point on the parabola to the focus is equal to the perpendicular distance from that point to the directrix. Since the directrix is the line x = -f, the distance from a point (x, y) on the parabola to the directrix is |x + f|.
By using the distance formula, we can express this condition mathematically:
x2+y2=|x+f|
Squaring both sides of the equation, we get:
x2+y2=(x+f)2
Expanding the right side of the equation:
x2+y2=x2+2xf+f2
Simplifying the equation, we find:
y2=2xf+f2
Now, we can differentiate both sides of the equation with respect to x to obtain the differential equation:
ddx(y2)=ddx(2xf+f2)
Using the chain rule and differentiating term by term, we have:
2ydydx=2f+0
Simplifying further:
ydydx=f
This is the differential equation for parabolas with the vertex and focus on the x-axis.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

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

Didn't find what you were looking for?