Find the equation of the ellipse whose center is at (0,0), one vertex is at (3,0) and one end of minor axis is at(0,2)

Let's start with the original equation. It is in the form y=a(x-h)^2+k, where (h,k) is the vertex. In this case, h= - 2 and k= 0. Thus, is it y=x^2 and we have done a horizontal translation 2 units to the left. I assume we are going to translate this 4 units to the right, then 2 units up, and then reflect it in the x-axis. Before we do the reflection, we will be 2 units to the right and 2 units up. This equation would be y=(x-2)^2+2. If we reflect this in the x-axis, would we get a downward facing parabola y=-(x-2)^2 -2?

Good luck,

Paul

If the quadratic equation has no roots, you cannot find the roots. The discriminate is negative, so if we attempt to use the quadratic equation, we get no roots.

For example, y=x^2 +0x + 3

a=1, b=0, c=3

(-b+/- sqrt(b^2 -4ac))/2a

Substituting in the numbers, we get

x= (-0 +/-sqrt(0^2 - 4(1)3))/2
x= (0 +/- sqrt (-12))/2

We cannot do the square root of -12. Therefore, there are no roots. This is the same as having no x-intercepts. The discriminant is b^2-4ac. In this case it is -12. Thus, there are no real roots.

However, you can still determine the maximum or minimum, the vertex, the axis of symmetry, the y-intercept and the stretch/compression. With this information you can graph the equation.

In this case, the y-intercept is 3, the vertex is at (0,3), the axis of symmetry is x=0, the minimum value of the function is 3.

Good luck.

Let me know if you have any questions.

Paul

Factor the expression,
x^2+10x+25=(x+5)^2
The minimum is the point ( x=-5, y=0)
The axis of symmetry is the vertical line with equation x=-5

The volume of a solid of revolution can be calculated by making use of some methods of integral calculus. For that you will need the exact equation of the the surface that limits the solid. Nothing you can do on a calculator.
However for the spheroids there exit formulas you can use on a calculator.

• For an ellipsoid

If a, b, c are the lengths of the semiaxes (length of semiaxis=half the length of the axis)
V=(4/3)PI*a*b*c

• For an oblate spheroid

An oblate spheroid is the solid of revolution formed by the rotation of an ellipse about its minor axis (here an image of the cross section from Wikipedia).



V=(4/3)PI*(a^2)*b

• For a prolate spheroid

A prolate spheroid is the solid of revolution formed by the rotation of an ellipse about its major axis

V=(4/3)PI*a*(b^2)

I am afraid I do not understand what you want exactly (ellipse, parallelepiped?). The area of an ellipse with semi-major axis a and semi-minor axis b is given by
Area=PI*a*b

By definition, the y-intercept is the value of y when x=0. Just set x=0 in the equation and solve for y.
Hence 3(0)+6y=90 or 6y=90. This gives y=90/6=15. The point of intercept of the y-axis is (0,15).
To get the x-intercept, set y= 0 in the equation and solve for x.
3x+6(0)=90 or 3x=90. This gives x=90/3=30.
The point of intercept of the x-axis is (30,0).

I am not quite sure how the major axis of your hyperbola is directed and i do not know if the lengths you give are measures of the major and minor axes or the measures of the semi-major and semi-minor axes. So I am giving you the equations and the graphs so that you can decide for yourself what is appropriate for your problem.

Major axis parallel to the X-axis
Equation and graph


Center is at x=1 and y=-2, semi-major axis length is a=6, and semi-minor axis length is b=12


Major axis is parallel to the y-axis
Equation


Center is at x=1 and y=-2, semi-major axis length is a=12, and semi-minor axis length is b=6.

I trust you can customize the equations to fit your need.

Hello,
Sorry to say it but the question does not make sense. A circle is a figure that is perfectly defined by the knowledge of its center (or centre depending of the English you use) and its radius.

An ellipse is perfectly defined if you know its two foci (plural of focus) and the length each of its two axes (major and minor).
Is it the perimeter, the area, the parameter p of the ellipse you are lookong for.

Hope it will help you focus on your real the question.(The pun is intended)

V (2,0) and Focus (2,2)
since the focus is 2 units above the vertex, the parabola opens upward

vertex (h,k)
a is the focal length (distance between the vertex and the focus)
a = 2
(2 units above the vertex)

(x-h)^2 = 4a (y-k)
(x-2)^2 = 4(2) (y-0)
x^2 - 4x + 4 = 8 (y)
x^2 - 4x + 4 = 8y
x^2 - 4x - 8y + 4 = 0

**Answer: "(x-2)^2 = 8 (y)" or in expanded form "x^2 - 4x - 8y +4"

I'm not sure what your ellipse looks like but the equation for the area of an ellipse is given by:

pi times the long axis times the short axis

pi = 3.1416
The length of the long axis is from the center of the ellipse to the outer edge of the ellipse
The length of the short axis is from the center of the ellipse to the outer edge of the ellipse

Hope this helps Loringh