Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.

Asked by Pragya Singh | 1 year ago |  86

##### Solution :-

We know that the vertex is (0, 0) and the parabola is symmetric about the y-axis.

The equation of the parabola is either of the from x= 4ay or x2 = -4ay.

Given that the parabola passes through point (5, 2), which lies in the first quadrant.

So, the equation of the parabola is of the form

x2 = 4ay, while point (5, 2) must satisfy the equation x2 = 4ay.

Then,

52 = 4a(2)

25 = 8a

a = $$\dfrac{25}{8}$$

Thus, the equation of the parabola is

x2 = 4$$(\dfrac{25}{8})y$$

x2 = $$\dfrac{25y}{2}$$

2x2 = 25y

The equation of the parabola is 2x2 = 25y

Answered by Abhisek | 1 year ago

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