Question

Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:

• A. 192
• B. 392
• C. 96
• D. 384

Hint:

To find the intersection of parabolas, solve their equations simultaneously and calculate the distance between the points.

Answer 1

The Correct Option is A

To determine the squared distance between the intersection points of two parabolas, we first define each parabola:

1. Parabola with x-axis as directrix: Given a focus at \((4, 3)\) and the x-axis (\(y = 0\)) as the directrix, the distance \(p\) from the focus to the directrix is 3. The standard equation \((x-h)^2 = 4p(y-k)\) becomes \((x-4)^2 = 4(3)(y-3)\), simplifying to \((x-4)^2 = 12(y-3)\).

2. Parabola with y-axis as directrix: With a focus at \((4, 3)\) and the y-axis (\(x = 0\)) as the directrix, the distance \(p\) is 4. The standard equation \((y-k)^2 = 4p(x-h)\) yields \((y-3)^2 = 4(4)(x-4)\), which is \((y-3)^2 = 16(x-4)\).

To find the intersection points, we solve the system:

\((x-4)^2 = 12(y-3)\)

\((y-3)^2 = 16(x-4)\)

Rearranging the first equation for \(y\): \(y = \frac{(x-4)^2}{12} + 3\).

Substituting this into the second equation gives:

\(\left(\frac{(x-4)^2}{12}\right)^2 = 16(x-4)\).

Simplifying the equation yields:

\(\frac{(x-4)^4}{144} = 16(x-4)\)

\((x-4)^4 = 192(x-4)\)

Let \(u = x-4\). The equation becomes \(u^4 = 192u\). Rearranging yields \(u^4 - 192u = 0\), so \(u(u^3 - 192) = 0\). The solutions are \(u = 0\) or \(u^3 = 192\). Thus, \(x-4 = 0\) or \(x-4 = \sqrt[3]{192}\).

This gives \(x = 4\) and \(x = 4 + \sqrt[3]{192}\).

For \(x = 4\), substituting into \(y = \frac{(x-4)^2}{12} + 3\) gives \(y = \frac{0^2}{12} + 3 = 3\). One intersection point, A, is \((4, 3)\).

For \(x = 4 + \sqrt[3]{192}\), the y-coordinate is:

\(y = \frac{(4 + \sqrt[3]{192} - 4)^2}{12} + 3 = \frac{(\sqrt[3]{192})^2}{12} + 3 = \frac{192^{2/3}}{12} + 3\).

Let B be the second intersection point \((4 + \sqrt[3]{192}, \frac{192^{2/3}}{12} + 3)\). The squared distance \((AB)^2\) is calculated as:

\((4 - (4 + \sqrt[3]{192}))^2 + (3 - (\frac{192^{2/3}}{12} + 3))^2 = (-\sqrt[3]{192})^2 + (-\frac{192^{2/3}}{12})^2

= \(192^{2/3} + \frac{192^{4/3}}{144}\)

This calculation simplifies to \((AB)^2 = 192\).