Find the standard form of the equation of the ellipse with the given characteristics and center at the origin.
Vertical major axis passes through the points $( 0,7 ) \text { and } ( 5,0 )$ .

A) $\frac { x ^ { 2 } } { 7 } + \frac { y ^ { 2 } } { 5 } = 1$
B) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 49 } = 1$
C) $\frac { x ^ { 2 } } { 49 } - \frac { y ^ { 2 } } { 25 } = 1$
D) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 49 } = 1$
E) $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 25 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. ​
Vertical major axis passes through the points $( 0,7 ) \text { and } ( 5,0 )$ .
​
A) $\frac { x ^ { 2 } } { 7 } + \frac { y ^ { 2 } } { 5 } = 1$
B) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 49 } = 1$
C) $\frac { x ^ { 2 } } { 49 } - \frac { y ^ { 2 } } { 25 } = 1$
D) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 49 } = 1$
E) $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 25 } = 1$

Quizplus · Accepted Answer

Since the ellipse has a vertical major axis, the equation will have the form:
$\frac{(y-k)^2}{b^2} + \frac{(x-h)^2}{a^2} = 1$
where (h,k) is the center of the ellipse, a is the length of the semi-major axis, and b is the length of the semi-minor axis.
The center is given as the origin, so we have:
$\frac{y^2}{b^2} + \frac{x^2}{a^2} = 1$
To find a and b, we need to use the information about the points the major axis passes through. The distance between these points is the length of the major axis, which is 16 (since the difference in y-coordinates is 8, and the axis is vertical). We also know that the semi-major axis is half the length of the major axis, so a=8.
To find b, we need to use the Pythagorean theorem with the distance between the two points as the hypotenuse, and half the length of the minor axis as one leg. We have:
$ (\frac{b}{2})^2+8^2=16^2$
$\frac{b^2}{4} = 240$
$b^2=960$
So our equation becomes:
$\frac{y^2}{960} + \frac{x^2}{64} = 1$
Multiplying by 960 to eliminate the fraction, we get:
$y^2 + \frac{960}{64}x^2 = 960$
Simplifying, we get:
$y^2 + 15x^2 = 960$
So the standard form of the equation is:
$15x^2 + y^2 = 960$.

Find the Standard Form of the Equation of the Ellipse (0,7) and (5,0)( 0,7 ) \text { and } ( 5,0 )(0,7) and (5,0)

Find the Standard Form of the Equation of the Ellipse $( 0,7 ) \text { and } ( 5,0 )$