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Exercise 10.2 · Q12

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

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Since the parabola is symmetric about the yy-axis and has its vertex at the origin, its equation must be of the form x2=4ayx^2 = 4ay. Substituting the point (5,2)(5, 2) gives a=258a = \frac{25}{8}, so the equation is x2=252yx^2 = \frac{25}{2}y.

The key to this problem is reading the conditions carefully and matching them to the standard forms of a parabola.

When a parabola has its vertex at (0,0)(0,0) and is symmetric about the yy-axis, its axis must be the yy-axis itself. That means the parabola opens either upward or downward. The standard equation for such a parabola is x2=4ayx^2 = 4ay, where aa is the focal length. If a>0a > 0, the parabola opens upward; if a<0a < 0, it opens downward.

We don't yet know which direction it opens — that will be decided by the given point.

  1. Set up the general form. Since the vertex is at the origin and the axis is the yy-axis, the equation is:

x2=4ayx^2 = 4ay

Here aa is an unknown constant (the distance from the vertex to the focus, with sign indicating direction).

  1. Use the given point to find aa. The parabola passes through (5,2)(5, 2). Substitute x=5x = 5 and y=2y = 2:

52=4a(2)5^2 = 4a(2)

25=8a25 = 8a

a=258a = \frac{25}{8}

Since aa is positive, the parabola opens upward.

  1. Write the final equation. Substitute a=258a = \frac{25}{8} back into x2=4ayx^2 = 4ay: x2=4⋅258⋅yx^2 = 4 \cdot \frac{25}{8} \cdot y …

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