Question

Find an equation of the parabola that is symmetric about y-axis has its vertex at the
origin and passes though the point (10, 4). Also show its focus and direction on the
graph

Answer 1

Given: vertex at the origin and passes though the point (10, 4).  

To find: Find an equation of the parabola ?

Solution:

• Now we have given that the parabola is symmetric about y-axis so the standard equation will be:

               x² = ± 4ay

• Now we have given that parabola passes through the point (10, 4) then the Parabola lies on positive y-axis, so:

               x² =  4ay

• Now the point should satisfy the equation of the parabola:

               10² =  4a(4)

               100 = 16a

               a = 100/16

               a = 25/4

• So the equation of parabola is x² = 25/4y

• Now we know that focus of upward opening parabola is:

               4p (y - k) = (x - h)²

               |p| = focal length = 25/16

               4 (25/16) (y - 0) = (x - 0)²

• So the focus of e is:

               (0, 0+p) = (0, 0+25/16) = (0, 25/16)

• So the focus is, (0, 25/16)

Answer:

         The equation of parabola is x² = 25/4y and the focus is given by, (0, 25/16).