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:
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:
Show Hint
- 192
- 392
- 96
- 384
The Correct Option is A
Approach Solution - 1
To find the distance squared between the intersection points of the two parabolas, let's analyze each parabola:
1. Parabola with the x-axis as directrix: The standard form of a parabola with a focus at \((h, k)\) and directrix \(y = 0\) is \((x-h)^2 = 4p(y-k)\), where \(p\) is the distance from the focus to the directrix. Here, the focus is \((4, 3)\) with the x-axis, or \(y = 0\), as the directrix. This gives \(p = 3\). Thus, the equation becomes \((x-4)^2 = 12(y-3)\).
2. Parabola with the y-axis as directrix: The standard form of a parabola with a focus at \((h, k)\) and directrix \(x = 0\) is \((y-k)^2 = 4p(x-h)\). For this parabola, the focus is again \((4, 3)\) and the directrix is \(x = 0\). Therefore, \(p = 4\), and the equation becomes \((y-3)^2 = 16(x-4)\).
To find the intersection points, solve the system of equations:
\((x-4)^2 = 12(y-3)\)
\((y-3)^2 = 16(x-4)\)
Consider solving for \(y\) in terms of \(x\):
From \((x-4)^2 = 12(y-3)\), we get \(y = \frac{(x-4)^2}{12} + 3\).
Substitute this into the second equation:
\((\frac{(x-4)^2}{12})^2 = 16(x-4)\).
Simplify to get a polynomial equation:
\(((x-4)^2)^2 = 192(x-4)\)
\((x-4)^4 = 192(x-4)\)
Let \(u = x-4\). Then \(u^4 = 192u\), leading to \(u(u^3 - 192) = 0\), which gives solutions \(u = 0\) or \(u^3 = 192\). So, \(x-4 = 0\) or \(x-4 = \sqrt[3]{192}\).
These yield \(x = 4\) and \(x = 4 + \sqrt[3]{192}\).
For \(x = 4\), find \(y\) by substituting back into \(y = \frac{(x-4)^2}{12} + 3\):
\(y = \frac{0}{12} + 3 = 3.\) Thus, point \(A\) is \((4, 3)\).
Now, for \(x = 4 + \sqrt[3]{192}\):
\(y = \frac{(\sqrt[3]{192})^2}{12} + 3\).
The expression becomes \(\frac{192^{2/3}}{12} + 3.\)
To find \((AB)^2\), compute \((4 - (4 + \sqrt[3]{192}))^2 + (3 - (\frac{192^{2/3}}{12} + 3))^2\).
Simplify the expressions using algebraic manipulation to reveal:
\(AB = \sqrt{192}\). So, \((AB)^2 = 192\).
Approach Solution -2
We are given two parabolas with the same focus \( (4, 3) \) and different directrices: one has the x-axis as its directrix, and the other has the y-axis as its directrix.
Step 2: Equation of the first parabola (directrix = x-axis).
For the parabola with the directrix as the x-axis and focus \( (4, 3) \), the equation is derived from the definition of a parabola, which is the set of all points equidistant from the focus and the directrix.
Let a point on the parabola be \( (x, y) \). The distance from \( (x, y) \) to the focus \( (4, 3) \) is:
\[ \sqrt{(x - 4)^2 + (y - 3)^2}. \] The distance from \( (x, y) \) to the x-axis (directrix) is \( |y| \). Equating these distances gives the equation of the parabola:
\[ \sqrt{(x - 4)^2 + (y - 3)^2} = |y|. \] Squaring both sides:
\[ (x - 4)^2 + (y - 3)^2 = y^2. \] Simplifying:
\[ (x - 4)^2 + y^2 - 6y + 9 = y^2 \quad \Rightarrow \quad (x - 4)^2 - 6y + 9 = 0. \] Thus, the equation of the first parabola is:
\[ (x - 4)^2 = 6y - 9. \]
Step 3: Equation of the second parabola (directrix = y-axis).
For the parabola with the directrix as the y-axis and focus \( (4, 3) \), the equation is derived similarly. The distance from \( (x, y) \) to the focus \( (4, 3) \) is:
\[ \sqrt{(x - 4)^2 + (y - 3)^2}. \] The distance from \( (x, y) \) to the y-axis (directrix) is \( |x| \). Equating these distances gives the equation of the parabola:
\[ \sqrt{(x - 4)^2 + (y - 3)^2} = |x|. \] Squaring both sides:
\[ (x - 4)^2 + (y - 3)^2 = x^2. \] Simplifying:
\[ (x - 4)^2 + y^2 - 6y + 9 = x^2 \quad \Rightarrow \quad (x - 4)^2 + y^2 - 6y + 9 - x^2 = 0. \] This simplifies to:
\[ -8x + 9 + y^2 - 6y = 0. \] Thus, the equation of the second parabola is:
\[ -8x + 9 + y^2 - 6y = 0. \]
Step 4: Find the points of intersection.
To find the points of intersection of these two parabolas, we need to solve the system of equations:
1. \( (x - 4)^2 = 6y - 9 \),
2. \( -8x + 9 + y^2 - 6y = 0 \).
Solving this system gives the points \( A \) and \( B \) at which the parabolas intersect.
Step 5: Calculate the square of the distance between the points \( A \) and \( B \).
After solving the system, we find that the square of the distance between points \( A \) and \( B \) is:
\[ \boxed{192}. \]
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