Question:medium

The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 20y$ to the end of its latus rectum is

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The area of the triangle formed by the vertex and the latus rectum is always $2a^2$.
Updated On: May 14, 2026
  • $100$ sq. units
  • $40$ sq. units
  • $20$ sq. units
  • $50$ sq. units
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
We need to find the coordinates of the vertex and the endpoints of the latus rectum for the given parabola, and then calculate the area of the triangle formed by these three points.
Step 2: Key Formula or Approach:
For a parabola $x^2 = 4ay$:
Vertex is $V(0, 0)$.
Focus is $S(0, a)$.
Length of latus rectum is $4a$.
Endpoints of latus rectum are $L(2a, a)$ and $L'(-2a, a)$.
Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$.
Step 3: Detailed Explanation:
The given equation of the parabola is $x^2 = 20y$.
Comparing this with the standard form $x^2 = 4ay$, we get:
$4a = 20 \Rightarrow a = 5$.
The vertex of the parabola is at the origin, $V(0, 0)$.
The focus is at $S(0, a) = (0, 5)$.
The latus rectum is a line segment passing through the focus and perpendicular to the axis of symmetry (y-axis). It lies on the line $y = 5$.
The length of the latus rectum is $4a = 20$.
The endpoints of the latus rectum are equidistant from the y-axis, at a distance of $2a = 10$.
So, the endpoints are $L(10, 5)$ and $L'(-10, 5)$.
We need the area of the triangle $V L L'$.
The base of this triangle is the length of the latus rectum, $LL' = 20$.
The height of the triangle corresponding to this base is the perpendicular distance from the vertex $V(0,0)$ to the line $LL'$ ($y=5$), which is equal to $a = 5$.
Area of triangle $VLL'$ $= \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times 20 \times 5 = 10 \times 5 = 50$ sq. units.
Step 4: Final Answer:
The area of the triangle is $50$ sq. units.
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