Key Concept: The Principle of Concurrency in Coordinate Geometry

In coordinate geometry, if a point lies on multiple curves, its coordinates must satisfy the equations of all those curves simultaneously. This fundamental principle is the cornerstone for solving problems involving intersections. When the points of intersection of two curves (say, Curve A and Curve B) also lie on a third curve (Curve C), it means that these specific intersection points are common to all three curves. Algebraically, this implies that the $(x, y)$ coordinates of these points must satisfy Equation A, Equation B, and Equation C simultaneously. Our strategy will be to use algebraic substitution to find expressions for $x$ and $y$ (or $x^2$ and $y^2$) from two of the equations, and then substitute these expressions into the third equation to find any unknown parameters.

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Step-by-Step Derivations

1. Understand the Given Equations and Conditions

We are provided with three equations representing different geometric curves and a crucial condition on a parameter $b$:

• Equation of the Ellipse: $\frac{x^2}{16} + \frac{y^2}{b^2} = 1 \quad \text{(Equation 1)}$ This is an ellipse centered at the origin. The semi-major axis is $a=4$ (along the x-axis) and the semi-minor axis is $b$ (along the y-axis), or vice-versa, depending on the value of $b$.

• Equation of the Circle: $x^2 + y^2 = 4b \quad \text{(Equation 2)}$ This is a circle centered at the origin with a radius $R = \sqrt{4b} = 2\sqrt{b}$.

• Equation of the Auxiliary Curve: $y^2 = 3x^2 \quad \text{(Equation 3)}$ This equation can be rewritten as $y = \pm\sqrt{3}x$. These are two straight lines passing through the origin. These lines make angles of $60^\circ$ and $120^\circ$ with the positive x-axis, respectively.

We are also given a critical condition:

• $b > 4$

Our objective is to determine the specific value of $b$ that satisfies all these conditions.

2. Utilize the Condition of Concurrency at Intersection Points

The problem states that the points where the ellipse (Equation 1) and the circle (Equation 2) intersect, also lie on the curve $y^2 = 3x^2$ (Equation 3). This means that any such intersection point $(x, y)$ must simultaneously satisfy all three equations.

To find these common intersection points, we can strategically combine any two of the equations. It is often most efficient to start with the equations that are algebraically simpler to combine. In this case, Equation 2 (the circle) and Equation 3 (the auxiliary curve) are excellent choices because they directly involve $x^2$ and $y^2$ and are easy to substitute into each other.

• Substitute Equation 3 into Equation 2: We have the equation of the circle: $x^2 + y^2 = 4b$. And the equation of the auxiliary curve: $y^2 = 3x^2$. Let's substitute the expression for $y^2$ from Equation 3 into Equation 2: $x^2 + (3x^2) = 4b$ Why this step? By combining these two equations, we are finding the algebraic expressions for $x^2$ and $y^2$ that define the points lying on both the circle and the auxiliary curve. Since these are the same points that intersect the ellipse, their $x^2$ and $y^2$ values must be consistent across all three curves.

• Solve for $x^2$: $4x^2 = 4b$ $x^2 = b$ Why this is useful? We have now found an expression for the square of the x-coordinate of the intersection points in terms of $b$. Since $b > 4$, $x^2 = b$ will be positive, ensuring real values for $x$ (i.e., $x = \pm\sqrt{b}$).

• Solve for $y^2$: Now that we have $x^2 = b$, we can substitute this back into Equation 3 ($y^2 = 3x^2$) to find $y^2$: $y^2 = 3(b)$ $y^2 = 3b$ Why this is useful? Similarly, we have an expression for the square of the y-coordinate of the intersection points in terms of $b$. Since $b > 4$, $y^2 = 3b$ will also be positive, ensuring real values for $y$ (i.e., $y = \pm\sqrt{3b}$).

So, for the points of intersection that satisfy the circle and the auxiliary curve (and thus, all three curves), we have the relationships $x^2 = b$ and $y^2 = 3b$.

3. Substitute Derived Values into the Ellipse Equation

Since the points defined by $x^2 = b$ and $y^2 = 3b$ are the same intersection points that lie on the ellipse, these values must satisfy the ellipse's equation (Equation 1): $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$

• Substitute $x^2 = b$ and $y^2 = 3b$ into Equation 1: $\frac{b}{16} + \frac{3b}{b^2} = 1$ Why this step? This is the crucial link that brings the parameter $b$ (from the circle and auxiliary curve) into the ellipse's equation. By substituting, we form an equation that must hold true for the common points, and this equation will involve only $b$, allowing us to solve for it.

• Simplify the equation: We can simplify the second term, $\frac{3b}{b^2}$. Since we are given $b > 4$, $b$ is non-zero, so we can cancel one $b$ from the numerator and denominator: $\frac{b}{16} + \frac{3}{b} = 1$

4. Solve the Resulting Quadratic Equation for $b$}

Now we have a single algebraic equation with one unknown, $b$. Let's solve it:

• Eliminate denominators: To clear the denominators ($16$ and $b$), multiply every term in the entire equation by their Least Common Multiple (LCM), which is $16b$: $16b \left( \frac{b}{16} \right) + 16b \left( \frac{3}{b} \right) = 16b (1)$ This simplifies to: $b^2 + 48 = 16b$

• Rearrange into a standard quadratic equation: Move all terms to one side to form a quadratic equation of the form $Ab^2 + Bb + C = 0$: $b^2 - 16b + 48 = 0$

• Factor the quadratic equation: We need to find two numbers that multiply to $48$ (the constant term) and add up to $-16$ (the coefficient of $b$). These numbers are $-4$ and $-12$. $(b - 4)(b - 12) = 0$

• Find the possible values for $b$: Setting each factor to zero gives the potential solutions: $b - 4 = 0 \implies b = 4$ $b - 12 = 0 \implies b = 12$

5. Apply the Given Constraint to Determine the Unique Solution

The problem statement included a critical condition: $b > 4$. We must check our potential solutions against this constraint to identify the correct value of $b$.

• Test $b = 4$: This value does not satisfy the condition $b > 4$, because $4$ is not strictly greater than $4$. If $b=4$, the denominator $b^2$ in the ellipse equation would become $16$, making the ellipse $\frac{x^2}{16} + \frac{y^2}{16} = 1$, which is a circle $x^2+y^2=16$. Also, the original circle $x^2+y^2=4b$ would become $x^2+y^2=16$. In this specific case, the ellipse and the circle would be identical, and their intersection points would be all points on the circle. However, the condition $b>4$ explicitly excludes this case. Therefore, $b=4$ is an extraneous solution based on the problem's specific constraints.

• Test $b = 12$: This value does satisfy the condition $b > 4$, as $12$ is strictly greater than $4$.

Therefore, the only valid value for $b$ that satisfies all the given conditions is $12$.

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Tips for JEE Aspirants:

• Read Conditions Carefully: Always pay close attention to all given conditions, especially inequalities (like $b > 4$). They are crucial for filtering out extraneous solutions and selecting the correct answer from multiple possibilities.
• Systematic Substitution: When dealing with multiple equations, plan your substitutions. Look for the simplest equations to combine first to get expressions for $x, y, x^2,$ or $y^2$. This makes the process more efficient.
• Avoid Unnecessary Square Roots: Notice how we consistently worked with $x^2$ and $y^2$ throughout the problem. This is often more efficient and avoids introducing square roots and their associated complexities (like $\pm$ signs) until absolutely necessary. The final ellipse equation also uses $x^2$ and $y^2$, making this approach natural.
• Understand "Intersection Points Lie On": This phrase is a direct instruction that those specific points satisfy all relevant equations. It's the core of the problem-solving strategy, allowing you to link parameters between different curves.
• Check for Extraneous Solutions: Algebraic manipulation can sometimes introduce solutions that don't fit the original problem's domain or constraints. Always verify your final answers against all given information.

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Summary and Key Takeaway:

This problem is a classic application of the principle of concurrency in coordinate geometry. By recognizing that the points of intersection must satisfy all three given equations (ellipse, circle, and auxiliary curve), we systematically derived expressions for $x^2$ and $y^2$ in terms of $b$ from the circle and the auxiliary curve equations. These expressions were then substituted into the ellipse