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

Question

stuth1 · Accepted Answer

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

y^{2} = 4 a x

...(1) Differentiating both sides of equation with respect to x, we get

2 y (\frac{d y}{d x}) = 4 a x

Substituting the value of 4a from equation in (1)

y^{2} = (2 y (\frac{d y}{d x})) x

\to y^{2} = 2 x y (\frac{d y}{d x})

\to y^{2} - 2 x y (\frac{d y}{d x}) = -

which is the Differential equations of the family of parabolas.

nick1337 · Accepted Answer

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:

y^{2} = 4 p x

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

x - h

. This gives us:

y^{2} = 4 p (x - h)

Expanding the right side, we have:

y^{2} = 4 p x - 4 p h

Simplifying further, we obtain:

y^{2} - 4 p x + 4 p h = 0

Finally, we can write the differential equation for this parabola by differentiating both sides with respect to

x

:

\frac{d y}{d x} (y^{2}) - \frac{d y}{d x} (4 p x) + \frac{d y}{d x} (4 p h) = 0

Using the chain rule and simplifying, we get:

2 y \cdot \frac{d y}{d x} - 4 p + 0 = 0

Thus, the differential equation for parabolas with the vertex and focus on the x-axis is:

2 y \cdot \frac{d y}{d x} - 4 p = 0

where

p

represents the distance from the vertex to the focus.

Don Sumner · Accepted Answer

Answer:dydx=2ax+bExplanation: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=14aSubstituting this value into the general equation, we have:y=ax2+bx+14aTaking the derivative of both sides with respect to x, we get:dydx=2ax+bTherefore, the differential equation for a parabola with its vertex and focus on the x-axis is:dydx=2ax+b

Vasquez · Accepted Answer

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:

\sqrt{x^{2} + y^{2}} = | x + f |

Squaring both sides of the equation, we get:

x^{2} + y^{2} = (x + f)^{2}

Expanding the right side of the equation:

x^{2} + y^{2} = x^{2} + 2 x f + f^{2}

Simplifying the equation, we find:

y^{2} = 2 x f + f^{2}

Now, we can differentiate both sides of the equation with respect to x to obtain the differential equation:

\frac{d}{d x} (y^{2}) = \frac{d}{d x} (2 x f + f^{2})

Using the chain rule and differentiating term by term, we have:

2 y \frac{d y}{d x} = 2 f + 0

Simplifying further:

y \frac{d y}{d x} = f

This is the differential equation for parabolas with the vertex and focus on the x-axis.

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

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