Find an equation of the parabola that is symmetric about y-axis has it

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

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

Answer»

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

10² = 4a(4)

100 = 16A

a = 100/16

a = 25/4

4p (y - k) = (x - h)²

|p| = FOCAL length = 25/16

4 (25/16) (y - 0) = (x - 0)²

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

Answer:

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