The given parabola is \(x ^2 = 12y.\)
On comparing this equation with \(x^ 2 = 4ay\), we obtain
\(4a = 12\)
\(⇒ a = 3\)
∴The coordinates of foci are \(S (0, a) = S (0, 3)\)
Let AB be the latus rectum of the given parabola.
The given parabola can be roughly drawn as
At \(y = 3, x^2 = 12 (3)\)
\(⇒ x^2 = 36\)
\(⇒ x = ±6\)
∴The coordinates of A are (-6, 3), while the coordinates of B are (6, 3).
Therefore, the vertices of ΔOAB are O (0, 0), A (-6, 3), and B (6, 3).
Area of\(\triangle OAB = \frac{1}{2} [0(3-3) + (-6)(3-0) + 6(0-3)]\) unit2
\(= \frac{1}{2} [(-6) (3) + 6 (-3)]\) unit2
\(= \frac{1}{2} [-18-18]\) unit2
\(= \frac{1}{2}[-36]\) unit2
\(= 18\) unit2
Thus, the required area of the triangle is 18 unit2 .
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
=> MP2 = PS2
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2