 Hello and welcome to the session let us discuss the following question it says find the area of the triangle formed by the lines joining the vertex of the parabola x square is equal to 12y to the n's of its latest rectum. So here we have to find the area of the triangle joining the vertex of the parabola to the n's of its latest rectum and latest rectum is the line passing through the focus so let us now move on to the solution we know that the general equation of parabola having symmetry of x's as y axis is given by x square is equal to 4a y let us call this as 1 and the given equation parabola is x square is equal to 12y since the equation involves x square so the symmetry of x's or x's of symmetry is the y axis let us call this as 2 and comparing 1 and 2 we have x square is equal to 4 into 3y so therefore a is 3 so focus is 0 3 and the length of latest rectum is equal to 4a now here a is 3 so this becomes 4 into 3 that is 12 now area of triangle o pq is equal to 1 by 2 into base into height now your base is pq and pq is the latest rectum which has length 12 units and height is a because the latest rectum passes through the point 0 a so here height is a units and a is equal to 3 so base is pq which is 12 units and height is given by a which is 3 units so area of triangle o pq is equal to 1 by 2 into 12 into 3 square units which is equal to 18 square units hence the area of the triangle is 18 square units this completes the question and the session by for now take care have a good day