Answer :
The standard form of the equation of the ellipse [tex]x^2 / 16 + y^2 / 3 = 1[/tex].
What is standard form of the equation of the ellipse?
For example, d2ydx2+4y=0 if your differential equation is homogeneous (it equals zero and not any function). y(x=0)=0 and y(x=2)=0 were given as the boundary conditions, and you were requested to answer the equation. Then the aforementioned boundary conditions are homogenous boundary conditions.
An ellipse is a type of conic section that is defined as the set of all points such that the sum of the distances from the point to two fixed points (called the foci) is constant. The standard form of the equation of an ellipse with center (h, k) is:
[tex](x - h)^2 / a^2 + (y - k)^2 / b^2 = 1[/tex]
where a and b are the lengths of the major and minor axes, respectively, and (h, k) is the coordinates of the center of the ellipse.
In this case, the center of the ellipse is (0, 0) and the focus is (3, 0), so we can use the distance formula to find the distance between the center and focus:
[tex]\sqrt{((3 - 0)^2 + (0 - 0)^2)} = 3[/tex]
The distance between the center and focus is equal to the length of the major axis, so a = 3.
In addition, the vertex of the ellipse is (4, 0), which is 4 units away from the center. Since the distance between the center and vertex is equal to the length of the major axis, we can set a = 4 to find the length of the minor axis:
[tex](x - 0)^2 / 4^2 + (y - 0)^2 / b^2 = 1[/tex]
Solving for b, we find that b = [tex]\sqrt{(3)}.[/tex]
Therefore, the standard form of the equation of the ellipse with the given characteristics is:
[tex](x - 0)^2 / 4^2 + (y - 0)^2 / \sqrt{(3)^2} = 1[/tex]
This can be simplified to:
[tex]x^2 / 16 + y^2 / 3 = 1[/tex]
The standard form of the equation of the ellipse [tex]x^2 / 16 + y^2 / 3 = 1[/tex].
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