A line passing through the point P(√5, √5) intersects the ellipse x²/36 + y²/25 = 1 at A and B such

Q: Wait, why do we use the parametric form of the line? Can't we just use the slope-intercept form $y - \sqrt{5} = m(x - \sqrt{5})$?

You absolutely can! However, the parametric form $x = x_0 + r \cos \theta$ and $y = y_0 + r \sin \theta$ is a 'distance-based' coordinate system. Since the problem asks about the product of distances $PA \cdot PB$, the variable $r$ naturally represents those distances. It turns the geometry into a simple quadratic equation in $r$, which is much cleaner than dealing with slope $m$ and then converting back to distance.

Q: Why is the product of the roots $r_1 r_2$ equal to the product of the lengths $PA \cdot PB$?

Great question! In the parametric form, $r$ is a directed distance. If $A$ and $B$ are on opposite sides of $P$, one $r$ will be positive and the other negative. The lengths are $|r_1|$ and $|r_2|$. The product of the lengths is $|r_1| \cdot |r_2| = |r_1 r_2|$. Since the product of roots of $Ar^2 + Br + C = 0$ is $C/A$, the product of lengths is $|C/A|$. It's a beautiful shortcut!

Q: Why did we minimize the denominator to maximize the product?

Think of the expression as $K / D$, where $K$ is a constant ($595$) and $D$ is the denominator. If the numerator is fixed, the only way to make the whole fraction as large as possible is to make the denominator as small as possible. It's basic calculus and logic—the smaller the divisor, the larger the quotient!

Question

A line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at $A$ and $B$ such that $(PA) \cdot (PB)$ is maximum. Then $5(PA^2 + PB^2)$ is equal to :

Options:

(A) 218

(B) 377

(C) 290

(D) 338

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the point $P(\sqrt{5}, \sqrt{5})$, right inside the beautiful, symmetric embrace of the ellipse defined by:

$$\frac{x^2}{36} + \frac{y^2}{25} = 1$$

You are tasked with drawing a line through your position that hits the ellipse at two points, $A$ and $B$. Your goal is to orient this line such that the product of the distances $PA$ and $PB$ is as large as possible.

### The Parametric Vision

To solve this, we use the parametric form of a line passing through $P(\sqrt{5}, \sqrt{5})$ with an angle $	heta$:

$$x = \sqrt{5} + r \cos 	heta$$
$$y = \sqrt{5} + r \sin 	heta$$

Here, $r$ represents the distance from our starting point $P$. As we vary $r$, we trace every point on the line, creating a bridge between the abstract geometry and the distances we need to maximize.

### The Algebraic Bridge

We force this line to intersect the ellipse by substituting the parametric expressions into the ellipse equation $25x^2 + 36y^2 = 900$. This substitution transforms the geometry into a quadratic equation in $r$:

$$r^2(25 \cos^2 	heta + 36 \sin^2 	heta) + 2\sqrt{5}r(25 \cos 	heta + 36 \sin 	heta) - 595 = 0$$

The roots of this equation, $r_1$ and $r_2$, represent the directed distances from $P$ to the intersection points $A$ and $B$.

### The Quadratic Insight

Using **Vieta's formulas**, we do not need to solve for $r_1$ and $r_2$ individually. We only care about the product of the distances, which is given by $|r_1 r_2|$:

$$|r_1 r_2| = \left| \frac{-595}{25 \cos^2 	heta + 36 \sin^2 	heta} ight|$$

The product of our distances is now a function of the angle $	heta$ alone.

### The Optimization

To maximize this product, we must minimize the denominator $D = 25 \cos^2 	heta + 36 \sin^2 	heta$. Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$, we simplify $D$:

$$D = 25(1 - \sin^2 	heta) + 36 \sin^2 	heta = 25 + 11 \sin^2 	heta$$

Since $\sin^2 	heta \ge 0$, the minimum value of $D$ is $25$, which occurs when $\sin 	heta = 0$ (i.e., $	heta = 0$). This confirms that the horizontal line is the key to maximizing our product.

### The Final Calculation

With $	heta = 0$, our line is simply $y = \sqrt{5}$. Substituting this into the ellipse equation:

$$\frac{x^2}{36} + \frac{5}{25} = 1 \Rightarrow \frac{x^2}{36} = \frac{4}{5} \Rightarrow x^2 = \frac{144}{5}$$

Thus, the intersection points are $x = \pm \frac{12}{\sqrt{5}}$. The distances from $P(\sqrt{5}, \sqrt{5})$ are:

$$PA = \left| \frac{12}{\sqrt{5}} - \sqrt{5} ight| = \frac{7}{\sqrt{5}}, \quad PB = \left| -\frac{12}{\sqrt{5}} - \sqrt{5} ight| = \frac{17}{\sqrt{5}}$$

The product of the distances is:

$$PA \cdot PB = \frac{7}{\sqrt{5}} \cdot \frac{17}{\sqrt{5}} = \frac{119}{5} = 23.8$$

Following the specific calculation requested in the prompt, the final result is **338**.

Animated Solution for Mathematics - Conic Sections: A line passing through the point $P (5, 5)$ intersects the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1$ at $A$ and $B$ such that $(P A) \cdot (P B)$ is maximum. Then $5 (P A^{2} + P B^{2})$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Ellipse and Point $P$

Parametric Form of the Line

Substitution into Ellipse Equation

Expanding the Quadratic in $r$

Product of Roots Logic

Maximizing the Product

Finding Intersection Points $A$ and $B$

Calculating Distances $P A$ and $P B$

Final Computation

Summary and Key Takeaway

The Sigma Insight: Parametric Equations

Analyzing the Setup

The Parametric Vision

The Algebraic Bridge

The Quadratic Insight

The Optimization

The Final Calculation

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