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).(PB) is maximum. Then 5(PA$^2$ + PB$^2$) is equal to :

• A. 218
• B. 377
• C. 290
• D. 338

Hint:

To maximize the product of two distances from a point to the intersection points of a line and an ellipse, the line should be parallel to the major or minor axis of the ellipse.

Answer 1

The Correct Option is D

Given ellipse is \[ \frac{x^2}{36} + \frac{y^2}{25} = 1 \] Any point on line \(AB\) can be assumed as \[ Q(\sqrt{5} + r\cos\theta, \sqrt{5} + r\sin\theta) \] Substituting this into the equation of the ellipse: \[ 25(\sqrt{5} + r\cos\theta)^2 + 36(\sqrt{5} + r\sin\theta)^2 = 900 \] Expanding and simplifying: \[ r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) + 25 \cdot 5 + 36 \cdot 5 = 900 \] \[ r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) = 900 - 305 = 595 \] \[ \Rightarrow r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) - 595 = 0 \] Let the roots of this quadratic in \(r\) be \(PA\) and \(PB\), then \[ PA \cdot PB = \frac{595}{25\cos^2\theta + 36\sin^2\theta} \] To maximize \(PA \cdot PB\), the denominator should be minimized: \[ 25\cos^2\theta + 36\sin^2\theta = 25 + 11\sin^2\theta \] Maximum value of \(PA \cdot PB\) occurs when \(\sin^2\theta = 0\), i.e., \(\theta = 0\) or \(\pi\) \[ \Rightarrow \text{Line } AB \text{ must be parallel to the x-axis} \Rightarrow y_A = y_B = \sqrt{5} \] Putting \(y = \sqrt{5}\) in the equation of the ellipse: \[ \frac{x^2}{36} + \frac{5}{25} = 1 \Rightarrow \frac{x^2}{36} = \frac{4}{5} \Rightarrow x^2 = \frac{4}{5} \cdot 36 = \frac{144}{5} \] Hence, coordinates of \(A\) and \(B\) are: \[ A = \left(-\frac{12}{\sqrt{5}}, \sqrt{5}\right), \quad B = \left(\frac{12}{\sqrt{5}}, \sqrt{5} \right) \] Now, \[ PA^2 + PB^2 = \left(\sqrt{5} - \frac{12}{\sqrt{5}}\right)^2 + \left(\sqrt{5} + \frac{12}{\sqrt{5}}\right)^2 \] \[ = 2\left(5 + \frac{144}{5} \right) = 2 \cdot \frac{169}{5} = \frac{338}{5} \] \[ \Rightarrow 5(PA^2 + PB^2) = 338 \]

Answer 2

The problem asks for the value of \( 5(\text{PA}^2 + \text{PB}^2) \) where A and B are the intersection points of a line passing through P\((\sqrt{5}, \sqrt{5})\) with the ellipse \( \frac{x^2}{36} + \frac{y^2}{25} = 1 \). The line is oriented such that the product of the distances, (PA)(PB), is maximized.

Concept Used:

To solve this problem, we use the parametric equation of a line passing through a given point. The intersection of this line with the ellipse results in a quadratic equation in terms of the distance parameter \(r\). The roots of this equation, \(r_1\) and \(r_2\), represent the distances PA and PB.

1. Parametric equation of a line: A line passing through \( P(x_1, y_1) \) can be represented as \( x = x_1 + r \cos \theta, y = y_1 + r \sin \theta \), where \(r\) is the distance from P to any point \((x, y)\) on the line, and \(\theta\) is the angle the line makes with the positive x-axis.

2. Intersection with Ellipse: Substituting these parametric coordinates into the ellipse equation yields a quadratic equation in \(r\), of the form \( Ar^2 + Br + C = 0 \).

3. Product of Roots: The product of the roots of this quadratic equation is \( r_1 r_2 = \frac{C}{A} \). Since the point P is inside the ellipse, the points A and B are on opposite sides of P along the line. Thus, one root will be positive and the other negative. The product of distances is \( \text{PA} \cdot \text{PB} = |r_1 r_2| = \left| \frac{C}{A} \right| \).

4. Maximizing the Product: To maximize the product of distances, we need to find the value of \(\theta\) that optimizes the expression \( \left| \frac{C}{A} \right| \).

Step-by-Step Solution:

Step 1: Set up the parametric equation of the line passing through P\((\sqrt{5}, \sqrt{5})\).

\[ x = \sqrt{5} + r \cos \theta, \quad y = \sqrt{5} + r \sin \theta \]

Step 2: Substitute these coordinates into the ellipse equation \( \frac{x^2}{36} + \frac{y^2}{25} = 1 \).

\[ \frac{(\sqrt{5} + r \cos \theta)^2}{36} + \frac{(\sqrt{5} + r \sin \theta)^2}{25} = 1 \]

Multiplying by the LCM, which is \(36 \times 25 = 900\), we get:

\[ 25(\sqrt{5} + r \cos \theta)^2 + 36(\sqrt{5} + r \sin \theta)^2 = 900 \]

Step 3: Expand and rearrange the equation to form a quadratic in \(r\).

\[ 25(5 + 2\sqrt{5}r \cos \theta + r^2 \cos^2 \theta) + 36(5 + 2\sqrt{5}r \sin \theta + r^2 \sin^2 \theta) = 900 \] \[ (125 + 50\sqrt{5}r \cos \theta + 25r^2 \cos^2 \theta) + (180 + 72\sqrt{5}r \sin \theta + 36r^2 \sin^2 \theta) = 900 \]

Group the terms by powers of \(r\):

\[ r^2(25 \cos^2 \theta + 36 \sin^2 \theta) + r(50\sqrt{5} \cos \theta + 72\sqrt{5} \sin \theta) + (125 + 180 - 900) = 0 \] \[ r^2(25 \cos^2 \theta + 36 \sin^2 \theta) + 2\sqrt{5}r(25 \cos \theta + 36 \sin \theta) - 595 = 0 \]

Step 4: Find the expression for the product of distances (PA) \(\cdot\) (PB).

The product of the roots \( r_1 r_2 \) is \( \frac{C}{A} \). The product of the distances is \( |\frac{C}{A}| \).

\[ \text{PA} \cdot \text{PB} = \left| \frac{-595}{25 \cos^2 \theta + 36 \sin^2 \theta} \right| = \frac{595}{25 \cos^2 \theta + 36 \sin^2 \theta} \]

Step 5: Maximize the product (PA) \(\cdot\) (PB).

To maximize this expression, we need to minimize the denominator. Let's rewrite the denominator:

\[ D(\theta) = 25 \cos^2 \theta + 36 \sin^2 \theta = 25(1 - \sin^2 \theta) + 36 \sin^2 \theta = 25 + 11 \sin^2 \theta \]

The minimum value of \( D(\theta) \) occurs when \( \sin^2 \theta \) is minimum, which is \( \sin^2 \theta = 0 \). This happens when \( \theta = 0 \) or \( \pi \), meaning the line is horizontal.

Step 6: Find the intersection points A and B for the horizontal line \( y = \sqrt{5} \).

Substitute \( y = \sqrt{5} \) into the ellipse equation:

\[ \frac{x^2}{36} + \frac{(\sqrt{5})^2}{25} = 1 \implies \frac{x^2}{36} + \frac{5}{25} = 1 \implies \frac{x^2}{36} = 1 - \frac{1}{5} = \frac{4}{5} \] \[ x^2 = \frac{144}{5} \implies x = \pm \frac{12}{\sqrt{5}} = \pm \frac{12\sqrt{5}}{5} \]

The intersection points are \( A\left(\frac{12\sqrt{5}}{5}, \sqrt{5}\right) \) and \( B\left(-\frac{12\sqrt{5}}{5}, \sqrt{5}\right) \).

Step 7: Calculate the distances PA and PB.

The point is P\((\sqrt{5}, \sqrt{5})\).

\[ \text{PA} = \left| \frac{12\sqrt{5}}{5} - \sqrt{5} \right| = \left| \frac{12\sqrt{5} - 5\sqrt{5}}{5} \right| = \frac{7\sqrt{5}}{5} \] \[ \text{PB} = \left| -\frac{12\sqrt{5}}{5} - \sqrt{5} \right| = \left| \frac{-12\sqrt{5} - 5\sqrt{5}}{5} \right| = \frac{17\sqrt{5}}{5} \]

Final Computation & Result:

Now, we compute the required expression \( 5(\text{PA}^2 + \text{PB}^2) \).

\[ \text{PA}^2 = \left(\frac{7\sqrt{5}}{5}\right)^2 = \frac{49 \times 5}{25} = \frac{49}{5} \] \[ \text{PB}^2 = \left(\frac{17\sqrt{5}}{5}\right)^2 = \frac{289 \times 5}{25} = \frac{289}{5} \] \[ \text{PA}^2 + \text{PB}^2 = \frac{49}{5} + \frac{289}{5} = \frac{338}{5} \]

Finally, we multiply by 5:

\[ 5(\text{PA}^2 + \text{PB}^2) = 5 \times \frac{338}{5} = 338 \]

The value of \( 5(\text{PA}^2 + \text{PB}^2) \) is 338.