Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.
Solution and Explanation
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 .
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.