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Miscellaneous Exercise · Q6

Q.Find the area of the triangle formed by the lines joining the vertex of the parabola x2=12yx^2 = 12y to the ends of its latus rectum.

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The vertex of the parabola x2=12yx^2 = 12y is at the origin, and its latus rectum is the horizontal line y=3y = 3 with endpoints (−6,3)(-6, 3) and (6,3)(6, 3). The triangle formed by the vertex and these two endpoints has base length 1212 and height 33, so its area is 12×12×3=18\frac12 \times 12 \times 3 = 18 square units.


1. Understanding the parabola

The given equation is x2=12yx^2 = 12y. This is a standard upward-opening parabola of the form x2=4ayx^2 = 4ay, where 4a=124a = 12, so a=3a = 3.

For a parabola x2=4ayx^2 = 4ay:

  • Vertex is at (0,0)(0, 0).
  • Focus is at (0,a)=(0,3)(0, a) = (0, 3).
  • Latus rectum is the line through the focus perpendicular to the axis (the yy-axis), i.e., the horizontal line y=a=3y = a = 3.

The latus rectum is a chord of the parabola passing through the focus and parallel to the directrix. Its endpoints lie on the parabola itself.

Tip

For x2=4ayx^2 = 4ay, the length of the latus rectum is always 4a4a. Here 4a=124a = 12, so the latus rectum is 1212 units long. This is a quick check: the endpoints will be symmetric about the yy-axis, each at x=±2a=±6x = \pm 2a = \pm 6.


2. Finding the endpoints of the latus rectum

The latus rectum lies on y=3y = 3. Substitute into x2=12yx^2 = 12y:

x2=12×3=36⇒x=±6.x^2 = 12 \times 3 = 36 \quad\Rightarrow\quad x = \pm 6.

So the endpoints are (−6,3)(-6, 3) and (6,3)(6, 3).


3. The triangle formed

The problem asks for the triangle formed by joining the vertex (0,0)(0, 0) to the ends of the latus rectum (−6,3)(-6, 3) and (6,3)(6, 3).

Plot these three points:

  • Vertex V=(0,0)V = (0, 0)
  • Left endpoint L=(−6,3)L = (-6, 3)
  • Right endpoint R=(6,3)R = (6, 3) …

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