[-/1 Points] DETAILS MY NOTES LARPCALC10 10.3.030.MI. Find the standard form of the equation of the ellipse with the given characteristics. Vertices: \( (4,0),(4,10) \); endpoints of the minor axis: \( (1,5),(7,5) \) Need Help? Read It Watch It Master It SUBMIT ANSWER

To find the standard form of the ellipse based on the given vertices and endpoints of the minor axis, let's begin by analyzing the information. The vertices \( (4,0) \) and \( (4,10) \) indicate that the major axis is vertical because their x-coordinates are the same. The center of the ellipse is the midpoint of the vertices, which is \( (4,5) \). Next, we can determine the lengths of the semi-major and semi-minor axes. The distance between the vertices is 10, thus the semi-major axis \( a \) is 5. The endpoints of the minor axis \( (1,5) \) and \( (7,5) \) show that the minor axis is horizontal, with a length of 6, giving us a semi-minor axis \( b \) of 3. This means the standard form for a vertically aligned ellipse centered at \( (h,k) \) is given by: \[ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \] Substituting your values \( h = 4 \), \( k = 5 \), \( a = 5 \), and \( b = 3 \), the equation becomes: \[ \frac{(x-4)^2}{3^2} + \frac{(y-5)^2}{5^2} = 1 \] \[ \frac{(x-4)^2}{9} + \frac{(y-5)^2}{25} = 1 \] So, the standard form of the ellipse equation is: \[ \frac{(x-4)^2}{9} + \frac{(y-5)^2}{25} = 1 \]