The area of the triangle formed by the lines | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation of the parabola is $x^{2}=12y$. Comparing this with the standard form $x^{2}=4ay$,we get $4a=12$,which implies $a=3$. The verte

Vedclass · Accepted Answer

$18 	ext{ sq. units}$

Vedclass · Answer

(A) The given equation of the parabola is $x^{2}=12y$. Comparing this with the standard form $x^{2}=4ay$,we get $4a=12$,which implies $a=3$. The vertex of the parabola is at $(0,0)$. The ends of the latus rectum are at $(2a, a)$ and $(-2a, a)$,which are $(6, 3)$ and $(-6, 3)$. The area of the triangle formed by the vertex $(0,0)$ and the points $(6, 3)$ and $(-6, 3)$ is given by: $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$ Here,the base is the length of the latus rectum,which is $4a = 12$. The height is the distance from the vertex to the latus rectum,which is $a = 3$. $	ext{Area} = \frac{1}{2} 	imes 12 	imes 3 = 18 	ext{ sq. units}$.

The area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12y$ to the ends of the latus rectum is

Similar Questions

If the tangent to the parabola $y^2 = ax$ makes an angle of $45^{\circ}$ with the $x$-axis,then the point of contact is

$A$ normal chord of the parabola $y^2 = 4x$ subtending a right angle at the vertex makes an acute angle $\theta$ with the $x$-axis. Then $\theta$ is equal to:

The points of intersection of the parabolas $y^2=5x$ and $x^2=5y$ lie on the line

If the parabola $x^{2} = 4ay$ passes through the point $(2, 1)$,then the length of the latus rectum is

The equation of any normal to the parabola ${y^2} = 4a(x - a)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module