Question

The focal points of an ellipse are (12, 0) and (−12, 0), and the point (12, 7) is on the ellipse. Find the points where this curve intersects the coordinate axes.

Answer 1

Answer:

x - ( 16 , 0 ) , ( -16 , 0 )

y - ( 0 ,  4√7 ) , ( 0 , -4√7)

Step-by-step explanation:

Solution:

- The sum of the distances from a point on the ellipse to its foci is constant. You have both foci and a point, so you can find the sum of the distances.

-Then you can find the vertices since they are points on the ellipse on the x-axis whose sum of distances to the foci are that value.

- The 7 in y coordinate of (12,7) is the length of semi-latus rectum. Also c is 12:

                                     c^2 = a^2 + b^2

Where, a: x-intercept

            b: y-intercept

- The length of semi-latus rectum is given by:

                                     b^2 = 7*a

- Substitute latus rectum expression in the first one we get:

                                    c^2 = a^2 + 7a

                                    a^2 + 7a - 144 = 0

                                    ( a - 16 ) * ( a - 9 ) = 0

                                    a = +/- ( 16 )

- The y-intercept we will use latus rectum expression again:

                                    b = +/- √(7*16)

                                    b = +/- 4√7

- The intercepts are:

                                    x - ( 16 , 0 ) , ( -16 , 0 )

                                    y - ( 0 ,  4√7 ) , ( 0 , -4√7)