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.
Answers
Given:
The parabola that is symmetric about y-axis has its vertex at the origin and passes though the point (10, 4).
To find:
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.
Solution:
From given, we have,
The parabola is symmetric about y-axis
Therefore, the standard equation is given by,
x² = ± 4ay
Since the given parabola passes through the point (10, 4) then the Parabola lies on positive y-axis
∴ x² = 4ay
The point should satisfy the equation of the parabola
10² = 4a(4)
100 = 16a
a = 100/16
∴ a = 25/4
The equation of parabola is x² = 25/4y
The focus of upward opening parabola is given by,
4p (y - k) = (x - h)²
|p| = focal length = 25/16
4 (25/16) (y - 0) = (x - 0)²
The focus of the e is given by, (0, 0+p) = (0, 0+25/16) = (0, 25/16)
The focus is, (0, 25/16)
∴ The equation of parabola is x² = 25/4y and the focus is given by, (0, 25/16)
