If the area of the triangle whose one vertex is at the vertex of the parabola, y² + 4(x - a²) = 0 an

Q: Wait, why is the vertex at $(a^2, 0)$ and not $(0, a^2)$?

Great question! Remember the standard form of a parabola is $(y - k)^2 = 4A(x - h)$, where $(h, k)$ is the vertex. In our equation $y^2 = -4(x - a^2)$, we can see that $k = 0$ and $h = a^2$. Thus, the vertex is $(a^2, 0)$. Always be careful to identify the shift correctly!

Q: Why is the base of the triangle $4a$?

When we set $x = 0$ in the parabola equation, we get $y^2 = 4a^2$, which gives $y = \pm 2a$. The two intersection points are $(0, 2a)$ and $(0, -2a)$. The distance between these two points on the y-axis is $|2a - (-2a)| = |4a|$. Since we are dealing with area, we take the positive magnitude, which is $4a$.

Q: Why can't we just use the general formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$?

You absolutely can! The determinant formula $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ will give you the same result. However, using the base-height method is often faster and more intuitive when one side of the triangle lies on an axis. Both methods are valid, but efficiency is key in JEE!

Question

If the area of the triangle whose one vertex is at the vertex of the parabola, $y^2 + 4(x - a^2) = 0$ and the other two vertices are the points of intersection of the parabola and y-axis, is 250 sq. units then a value of 'a' is:

Options:

(A) $5\sqrt{5}$

(B) 5

(C) $(10)^{2/3}$

(D) $(5)^{1/3}$

Sigma Percentile · Accepted Answer

### The Geometry of the Parabola: A Journey into Symmetry

Welcome, future engineer! Today, we are going to peel back the layers of a beautiful coordinate geometry problem. It is not just about plugging numbers into formulas; it is about visualizing the dance between algebraic equations and geometric shapes.

### Phase 1: Decoding the Equation

We start with the given equation: $y^2 + 4(x - a^2) = 0$.

By rearranging it, we get:

$$y^2 = -4(x - a^2)$$

This is the classic form of a parabola opening to the left. The term $(x - a^2)$ tells us that the entire shape has been shifted horizontally by $a^2$ units. This is our first anchor point.

### Phase 2: The Vertex and the Y-axis

Every parabola has a heart, its vertex. By comparing our equation to the standard form $Y^2 = -4AX$, we identify that the vertex $V$ is located at $(a^2, 0)$.

Now, imagine this parabola sitting on the coordinate plane, its vertex resting at $a^2$ on the x-axis, opening its arms towards the left. The problem asks us to consider the points where this curve intersects the y-axis.

To find these, we set $x = 0$. Substituting this into our equation:

$$y^2 = -4(0 - a^2)$$

$$y^2 = 4a^2$$

Taking the square root, we find $y = \pm 2a$. Thus, our parabola intersects the y-axis at two points: $P_1(0, 2a)$ and $P_2(0, -2a)$.

### Phase 3: Constructing the Triangle

Now, visualize the triangle formed by these three points: the vertex $V(a^2, 0)$ and the two intersection points $P_1(0, 2a)$ and $P_2(0, -2a)$.

This is a triangle with its base resting perfectly on the y-axis. The length of this base is the distance between $2a$ and $-2a$, which is $4a$.

The height of this triangle is the perpendicular distance from the vertex $V$ to the y-axis, which is simply the x-coordinate of the vertex, $a^2$.

### Phase 4: The Final Calculation

We are now ready to calculate the area. The formula for the area of a triangle is:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

Substituting our values, we get:

$$	ext{Area} = \frac{1}{2} 	imes (4a) 	imes (a^2) = 2a^3$$

The problem states that this area is $250$ square units. So, we set:

$$2a^3 = 250$$

Dividing by $2$, we get $a^3 = 125$. Taking the cube root of both sides, we arrive at our final answer:

**$a = 5$**

### Conclusion

See how elegant that was? By breaking the problem down into logical steps—identifying the vertex, finding the intercepts, and visualizing the geometry—we turned a daunting equation into a simple, solvable puzzle. Keep this mindset, and you will conquer any problem the JEE throws your way.

Animated Solution for Mathematics - Conic Sections: If the area of the triangle whose one vertex is at the vertex of the parabola, $y^{2} + 4 (x - a^{2}) = 0$ and the other two vertices are the points of intersection of the parabola and y-axis, is 250 sq. units then a value of 'a' is:

Select Answer:

Visualized Solution

Equation of the Parabola

Finding the Vertex

Y-axis Intersection Concept

Calculating Y-intercepts

Visualizing the Triangle

Determining Base and Height

Area Formula Setup

Simplifying Area Expression

Equating to Given Area

Solving for $a^{3}$

Final Value of $a$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Geometry of the Parabola

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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