Math Problem Statement

standard form of center (10,16) radius of elipse 3

Solution

To find the standard form of the equation of an ellipse with a given center and radius, we use the following general formula for an ellipse centered at (h,k)(h, k) with horizontal and vertical radii aa and bb:

(xh)2a2+(yk)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1

Given:

  • Center: (h,k)=(10,16)(h, k) = (10, 16)
  • Radius (assuming it is the same for both axes): a=b=3a = b = 3

Since the ellipse has equal radii along both axes, it is a circle. The equation of a circle is given by:

(x10)232+(y16)232=1\frac{(x - 10)^2}{3^2} + \frac{(y - 16)^2}{3^2} = 1

Simplifying, this becomes:

(x10)29+(y16)29=1\frac{(x - 10)^2}{9} + \frac{(y - 16)^2}{9} = 1

Or equivalently:

(x10)2+(y16)2=9(x - 10)^2 + (y - 16)^2 = 9

This is the standard form of the ellipse (which in this case is actually a circle).


Would you like more details or have any questions?

Here are some related questions you might be interested in:

  1. What if the radii along the two axes are different?
  2. How do you find the foci of an ellipse?
  3. How do you derive the standard form equation from the general equation of an ellipse?
  4. What is the significance of the major and minor axes in an ellipse?
  5. How do you calculate the area of an ellipse?
  6. How would the equation change if the ellipse was rotated?
  7. How do you find the eccentricity of an ellipse?
  8. How do you convert the standard form to the general form?

Tip: Remember, when the radii are equal, the ellipse becomes a circle, and the standard form simplifies accordingly.