If the points of intersection of two distinct conics x² + y² = 4b and x²/16 + y²/b² = 1 lie on the c

Q: Why can't we just solve for $x$ and $y$ directly from the start?

You could, but it would be an algebraic nightmare! By using the condition $y^2 = 3x^2$ as a bridge, we reduce the complexity. In JEE Advanced, the goal is often to find the most elegant path, not just the first one you see.

Q: Why do we reject $b=4$?

The problem explicitly states that the conics are 'distinct'. If $b=4$, the ellipse equation becomes $\frac{x^2}{16} + \frac{y^2}{16} = 1$, which simplifies to $x^2 + y^2 = 16$. This is identical to the circle $x^2 + y^2 = 4(4) = 16$. They are the same curve, not two distinct ones!

Q: How do we know the intersection points form a rectangle?

Because the equations $x^2 = b$ and $y^2 = 3b$ imply that for every $x$, there is a $\pm y$, and for every $y$, there is a $\pm x$. This symmetry about both the x-axis and y-axis forces the four intersection points to form a rectangle centered at the origin.

Question

If the points of intersection of two distinct conics $x^2 + y^2 = 4b$ and $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ lie on the curve $y^2 = 3x^2$, then $3\sqrt{3}$ times the area of the rectangle formed by the intersection points is ______

Sigma Percentile · Accepted Answer

### The Geometry of Intersection: A Masterclass in Symmetry

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a beautiful geometric puzzle. We are dealing with two conics—a circle and an ellipse—and a condition that binds them together.

When you see a problem like this, do not rush to the algebra. Pause. Breathe. Visualize.

### Phase 1: The Bridge of Symmetry

We are given a circle, $x^2 + y^2 = 4b$, and an ellipse,

$$ \frac{x^2}{16} + \frac{y^2}{b^2} = 1 $$

The problem states that their intersection points lie on the curve $y^2 = 3x^2$. This is our 'bridge'.

If you look closely at $y^2 = 3x^2$, you might be tempted to think of it as a parabola, but it is actually a pair of straight lines: $y = \sqrt{3}x$ and $y = -\sqrt{3}x$. These lines pass through the origin.

The fact that the intersection points lie on these lines tells us that the intersection points are symmetric about both the $x$-axis and the $y$-axis. This is a massive hint that our final shape will be a rectangle.

### Phase 2: The Algebraic Dance

Now, let us perform the substitution. We want to find the coordinates of these intersection points in terms of $b$.

We take our condition $y^2 = 3x^2$ and substitute it into the circle's equation:

$$ x^2 + (3x^2) = 4b $$

This simplifies beautifully to $4x^2 = 4b$, or simply $x^2 = b$. If $x^2 = b$, then $y^2$ must be $3b$.

We have now reduced the entire system to a single parameter, $b$. This is the power of substitution—we have turned a two-variable problem into a one-variable problem.

### Phase 3: The Trap of Identity

Next, we take our findings, $x^2 = b$ and $y^2 = 3b$, and substitute them into the ellipse equation:

$$ \frac{b}{16} + \frac{3b}{b^2} = 1 $$

Assuming $b 
eq 0$, this becomes:

$$ \frac{b}{16} + \frac{3}{b} = 1 $$

Multiplying by $16b$ gives us the quadratic $b^2 + 48 = 16b$, or $b^2 - 16b + 48 = 0$. Factoring this, we find $(b - 12)(b - 4) = 0$.

Here is where the JEE examiner tests your attention to detail. We have two candidates: $b = 12$ and $b = 4$.

If we choose $b = 4$, the ellipse becomes $\frac{x^2}{16} + \frac{y^2}{16} = 1$, which is $x^2 + y^2 = 16$. But our circle was $x^2 + y^2 = 4(4) = 16$.

They are the same! The problem demands 'distinct' conics, so we must reject $b = 4$. We proceed with $b = 12$.

### Phase 4: The Final Geometry

With $b = 12$, our intersection points are defined by $x^2 = 12$ and $y^2 = 3(12) = 36$. This gives us $x = \pm 2\sqrt{3}$ and $y = \pm 6$.

The vertices of our rectangle are $(2\sqrt{3}, 6)$, $(-2\sqrt{3}, 6)$, $(-2\sqrt{3}, -6)$, and $(2\sqrt{3}, -6)$.

The width is the distance between $2\sqrt{3}$ and $-2\sqrt{3}$, which is $4\sqrt{3}$. The height is the distance between $6$ and $-6$, which is $12$.

The area is simply:

$$ 4\sqrt{3} 	imes 12 = 48\sqrt{3} $$

Finally, the question asks for $3\sqrt{3}$ times this area:

$$ 3\sqrt{3} 	imes 48\sqrt{3} = 3 	imes 48 	imes 3 = \mathbf{432} $$

We have arrived at the solution, not by brute force, but by respecting the geometry and carefully navigating the constraints. Keep this mindset, and no problem will ever be too difficult.

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Using the Intersection Condition

Finding $x$ and $y$ in terms of $b$

Substituting into the Ellipse

Setting up the Equation for $b$

Solving the Quadratic for $b$

Checking the Distinct Conics Condition

Finding Intersection Points

Forming the Rectangle

Calculating the Area

Final Calculation

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

The Geometry of Intersection

Phase 1

Phase 2

Phase 3

Phase 4

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