Question:

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.

Updated On: Oct 24, 2023
Hide Solution
collegedunia
Verified By Collegedunia

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 

area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y

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 .

Was this answer helpful?
0
0

Concepts Used:

Parabola

Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).

Parabola


 

 

 

 

 

 

 

 

 

Standard Equation of a Parabola

For horizontal parabola

  • Let us consider
  • Origin (0,0) as the parabola's vertex A,
  1. Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
  2. 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
  1. Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
  2. 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.