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Exercise 10.2 · Q6

Q.Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum of the parabola x2=−9yx^2 = -9y.

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The parabola x2=−9yx^2 = -9y opens downward. Its focus is at (0,−94)(0, -\frac{9}{4}), the axis is the y-axis (x=0x = 0), the directrix is y=94y = \frac{9}{4}, and the length of the latus rectum is 99.


Concept and Intuition

The standard form for a vertical parabola is x2=4ayx^2 = 4ay.

  • If a>0a > 0, the parabola opens upward (focus above the vertex).
  • If a<0a < 0, it opens downward (focus below the vertex).

Here, the given equation is x2=−9yx^2 = -9y. Compare it with x2=4ayx^2 = 4ay:

4a=−9⇒a=−944a = -9 \quad \Rightarrow \quad a = -\frac{9}{4}

Since aa is negative, the parabola opens downward. The vertex is at the origin (0,0)(0,0).

For x2=4ayx^2 = 4ay:

  • Focus: (0,a)(0, a)
  • Directrix: y=−ay = -a
  • Axis: x=0x = 0 (the y-axis)
  • Length of latus rectum: ∣4a∣|4a|

Now let’s apply this step by step.


Step-by-step solution

1. Identify aa from the equation

We have x2=−9yx^2 = -9y. Writing it as x2=4ayx^2 = 4ay gives 4a=−94a = -9, so:

a=−94a = -\frac{9}{4}

2. Find the focus

For x2=4ayx^2 = 4ay, the focus is at (0,a)(0, a).

So focus = (0,−94)\left(0, -\frac{9}{4}\right).

Tip

The focus always lies on the axis of symmetry. Since the parabola opens downward, the focus is below the vertex — exactly ∣a∣|a| units away.

3. Find the axis

The axis is the line through the vertex perpendicular to the directrix. For a vertical parabola, it’s the y-axis:

Axis: x=0\text{Axis: } x = 0

4. Find the directrix

The directrix is a horizontal line opposite the focus, at y=−ay = -a.

Here −a=−(−94)=94-a = -\left(-\frac{9}{4}\right) = \frac{9}{4}, so: …

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