Animated Solution for Mathematics - Conic Sections: If the point of intersections of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the circle $x^2+y^2=4b,b>4$ lie on the curve $y^2=3x^2$, then $b$ is equal to:

Imagine you are standing in a coordinate plane, looking at three distinct entities: an ellipse, a circle, and a pair of lines. The ellipse is defined by:

The circle is defined by $x^2+y^2=4b$, and the lines are given by $y^2=3x^2$. These curves meet at common intersection points, which serves as our golden ticket to solving the system.

Instead of diving headfirst into the messy algebra of intersecting an ellipse and a circle, we use the lines as our bridge. Since the intersection points must satisfy all three equations, they must satisfy the line equation $y^2=3x^2$.

Substituting this into the circle equation $x^2+y^2=4b$:

This simplifies beautifully to $4x^2=4b$, or simply $x^2=b$. Consequently, the line equation $y^2=3x^2$ implies that $y^2=3b$. We have now expressed the coordinates of the intersection points entirely in terms of $b$.

Now, we bring the ellipse into the fold. Substituting $x^2=b$ and $y^2=3b$ into the ellipse equation $\frac{x^2}{16}+\frac{y^2}{b^2}=1$, we obtain:

Simplifying the second term, $\frac{3b}{b^2}$, we get:

To clear the denominators, we multiply the entire equation by $16b$, resulting in $b^2+48=16b$. Rearranging this yields the quadratic equation:

We factorize the quadratic equation $b^2-16b+48=0$ by looking for two numbers that multiply to $48$ and add to $-16$. These numbers are $-12$ and $-4$, leading to:

This gives us two potential values for $b$: $12$ and $4$. However, the problem explicitly states the constraint $b>4$.

We must reject $b=4$ because it violates the given condition. Therefore, the only valid solution is $b=12$.