Two parabolas have the same focus (4,3) and their directrices are the x-axis and the y-axis, respect

Q: Wait, why can we just equate the two equations? Doesn't that only give us the intersection?

Exactly! That is the point. We are looking for the intersection points $A$ and $B$. By equating the expressions for the distance to the focus, we find the locus of points that satisfy both conditions simultaneously. This is a powerful technique in coordinate geometry to find intersection points without solving complex systems.

Q: Why did we reject $y = -x$?

Look at the focus $(4, 3)$. It is in the first quadrant. Both parabolas open towards the first quadrant to reach that focus. The intersection must occur in the region where both parabolas exist, which is the first quadrant. The line $y = -x$ would imply points in the second or fourth quadrants, which does not fit the geometry of these specific parabolas.

Q: Do I really need to solve the quadratic equation $x^2 - 14x + 25 = 0$?

No! That is the beauty of using Vieta's formulas. We only need the value of $(x_1 - x_2)^2$. Using the identity $(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$ is much faster and less prone to calculation errors than finding the roots using the quadratic formula.

Question

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

Options:

(A) 392

(B) 384

(C) 192

(D) 96

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a coordinate plane. You have a fixed point, the focus $S(4, 3)$, and two lines, the x-axis ($y=0$) and the y-axis ($x=0$).

A parabola is the locus of all points $P(x, y)$ that maintain a perfect balance: the distance from $P$ to the focus $S$ must equal the perpendicular distance from $P$ to the directrix $L$. Mathematically, this is $PS = PM$, or $PS^2 = PM^2$.

For our first parabola, with the directrix $y=0$, the distance squared is:
$$(x-4)^2 + (y-3)^2 = y^2$$

For the second parabola, with the directrix $x=0$, the distance squared is:
$$(x-4)^2 + (y-3)^2 = x^2$$

### The Geometric Insight

We want to find where these two parabolas meet. At the intersection points $A$ and $B$, both equations must be true.

Notice that the left-hand side of both equations is identical: $(x-4)^2 + (y-3)^2$. This means we can set the right-hand sides equal to each other:
$$y^2 = x^2$$

This simplifies to $y = x$ or $y = -x$. Because our focus $(4, 3)$ is in the first quadrant, the parabolas must intersect in the first quadrant. Thus, we discard $y = -x$ and focus entirely on the line $y = x$.

### The Power of Algebra

Now, we substitute $y = x$ into our first equation:
$$(x-4)^2 + (x-3)^2 = x^2$$

Expanding this, we get:
$$(x^2 - 8x + 16) + (x^2 - 6x + 9) = x^2$$

Simplifying, we arrive at the quadratic equation:
$$x^2 - 14x + 25 = 0$$

Let the roots of this equation be $x_1$ and $x_2$. These are the x-coordinates of our intersection points $A$ and $B$. Since $y = x$, the points are $A(x_1, x_1)$ and $B(x_2, x_2)$.

We need $(AB)^2$. Using the distance formula:
$$(AB)^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$$

Since $y=x$, this becomes:
$$(AB)^2 = (x_1 - x_2)^2 + (x_1 - x_2)^2 = 2(x_1 - x_2)^2$$

### Final Calculation

We do not need to solve for $x_1$ and $x_2$ individually. We use the identity:
$$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$$

From our quadratic $x^2 - 14x + 25 = 0$, the sum of the roots is $x_1 + x_2 = 14$ and the product is $x_1x_2 = 25$. Thus:
$$(x_1 - x_2)^2 = 14^2 - 4(25) = 196 - 100 = 96$$

Finally, we calculate the squared distance:
$$(AB)^2 = 2 	imes 96 = 192$$

We have arrived at the answer with elegance and precision. The final result is **192**.

Animated Solution for Mathematics - Conic Sections: Two parabolas have the same focus $(4, 3)$ and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersects at the points A and B, then $(A B)^{2}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Setup

The Definition of a Parabola

Equation of Parabola 1

Equation of Parabola 2

Finding the Intersection

The Line of Intersection

Substituting $y = x$

Expanding the Equation

The Quadratic Equation

Distance AB Squared

Sum and Product of Roots

Calculating $(x_{1} - x_{2})^{2}$

Final Calculation

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

Analyzing the Setup

The Geometric Insight

The Power of Algebra

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Setup

The Definition of a Parabola

Equation of Parabola 1

Equation of Parabola 2

Finding the Intersection

The Line of Intersection

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x

Expanding the Equation

The Quadratic Equation

Distance AB Squared

Sum and Product of Roots

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x1​−x2​)2

Final Calculation

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

Analyzing the Setup

The Geometric Insight

The Power of Algebra

Final Calculation

Similar Questions

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