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Exercise 10.2 · Q9

Q.Find the equation of the parabola that satisfies the given conditions: Vertex (0,0)(0, 0); focus (3,0)(3, 0).

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The parabola has its vertex at the origin and focus on the positive x‑axis, so it opens to the right. The standard form is y2=4axy^2 = 4ax with a=3a = 3, giving the equation y2=12xy^2 = 12x.

A parabola is defined as the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix). When the vertex is at (0,0)(0,0) and the focus lies on the x‑axis at (3,0)(3,0), the axis of the parabola is the x‑axis itself. The vertex sits exactly midway between the focus and the directrix, so the directrix must be the vertical line x=−3x = -3.

Because the focus is to the right of the vertex, the parabola opens to the right. For a parabola with vertex at the origin and axis along the x‑axis, the standard equation is:

y2=4axy^2 = 4ax

Here aa is the signed distance from the vertex to the focus. Since the focus is at (3,0)(3,0), we have a=3a = 3. The directrix is x=−a=−3x = -a = -3.

Now let’s build the equation step by step.

  1. Identify the orientation.

    The focus (3,0)(3,0) is on the positive x‑axis and the vertex is at the origin. This tells us the parabola opens to the right. If it opened to the left, the focus would have a negative x‑coordinate.

  2. Recall the standard form for a right‑opening parabola.

    When the vertex is at (0,0)(0,0) and the axis is horizontal, the equation is y2=4axy^2 = 4ax. The parameter aa is the distance from the vertex to the focus (and also from the vertex to the directrix, but on the opposite side).

  3. Find aa from the focus.

    The focus is (a,0)(a, 0). We are given (3,0)(3,0), so a=3a = 3.

  4. Write the equation.

    Substitute a=3a = 3 into y2=4axy^2 = 4ax: …

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