Let's calculate T and S_1. The general form for T for an ellipse (x²)/(a²) + (y²)/(b²) = 1 at a point (x_1, y_1) is (xx_1)/(a²) + (yy_1)/(b²) = 1. So, for P(√5, √5), T is (x√5)/(36) + (y√5)/(25) - 1. The value S_1 is (√5)²36 + (√5)²25 - 1 = (5)/(36) + (5)/(25) - 1 = (5)/(36) + (1)/(5) - 1 = (25+36-180)/(180) = (61-180)/(180) = -(119)/(180).

Let's calculate T and S_1. The general form for T for an ellipse (x²)/(a²) + (y²)/(b²) = 1 at a point (x_1, y_1) is (xx_1)/(a²) + (yy_1)/(b²) = 1. So, for P(√5, √5), T is (x√5)/(36) + (y√5)/(25) - 1. The value S_1 is (√5)²36 + (√5)²25 - 1 = (5)/(36) + (5)/(25) - 1 = (5)/(36) + (1)/(5) - 1 = (25+36-180)/(180) = (61-180)/(180) = -(119)/(180).

Setting T=S_1, we get (x√5)/(36) + (y√5)/(25) - 1 = -(119)/(180). This simplifies to (x√5)/(36) + (y√5)/(25) = 1 - (119)/(180) = (180-119)/(180) = (61)/(180). Dividing by √5, we get (x)/(36) + (y)/(25) = (61)/(180√5). This is the equation of the chord AB.

Setting T=S_1, we get (x√5)/(36) + (y√5)/(25) - 1 = -(119)/(180). This simplifies to (x√5)/(36) + (y√5)/(25) = 1 - (119)/(180) = (180-119)/(180) = (61)/(180). Dividing by √5, we get (x)/(36) + (y)/(25) = (61)/(180√5). This is the equation of the chord AB.

We need to find 5(PA² + PB²). Since PA = PB = √((119)/(180)), we have PA² = PB² = (119)/(180). Substituting these values: 5((119)/(180) + (119)/(180)) = 5(2 · (119)/(180)) = 5((119)/(90)) = (595)/(90) = (119)/(18).

We need to find 5(PA² + PB²). Since PA = PB = √((119)/(180)), we have PA² = PB² = (119)/(180). Substituting these values: 5((119)/(180) + (119)/(180)) = 5(2 · (119)/(180)) = 5((119)/(90)) = (595)/(90) = (119)/(18).

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

Answer: $\frac{119}{18}$

Step-by-step solution

Step 1: Understand the problem and the power of a point theorem

The problem asks us to find $5(PA^2 + PB^2)$ for a line passing through point $P(\sqrt{5}, \sqrt{5})$ and intersecting the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at points $A$ and $B$ . The condition is that the product $PA \cdot PB$ is maximum. For a point $P$ outside an ellipse, the product $PA \cdot PB$ is constant for any line passing through $P$ and intersecting the ellipse. This constant value is equal to the square of the length of the tangent from $P$ to the ellipse, if a tangent exists. However, $P(\sqrt{5}, \sqrt{5})$ is inside the ellipse, as shown by substituting its coordinates into the ellipse equation: $\frac{(\sqrt{5})^2}{36} + \frac{(\sqrt{5})^2}{25} = \frac{5}{36} + \frac{5}{25} = \frac{1}{36/5} + \frac{1}{5} = \frac{1}{7.2} + \frac{1}{5} = 0.138 + 0.2 = 0.338 < 1$ . Therefore, the line segment $AB$ is a chord passing through $P$ . The product $PA \cdot PB$ is maximized when $P$ is the midpoint of the chord $AB$ . This occurs when the chord is perpendicular to the line connecting the center of the ellipse to $P$ .

Step 2: Find the equation of the chord AB

The center of the ellipse is $(0{,}0)$ . The point $P$ is $(\sqrt{5}, \sqrt{5})$ . The chord $AB$ for which $P$ is the midpoint is given by the equation $T=S_1$ , where $T$ is the tangent equation and $S_1$ is the value of the ellipse equation at point $P$ . The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{25} - 1 = 0$ .

Step 3: Calculate T and S1

Let's calculate $T$ and $S_1$ . The general form for $T$ for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$ . So, for $P(\sqrt{5}, \sqrt{5})$ , $T$ is $\frac{x\sqrt{5}}{36} + \frac{y\sqrt{5}}{25} - 1$ . The value $S_1$ is $\frac{(\sqrt{5})^2}{36} + \frac{(\sqrt{5})^2}{25} - 1 = \frac{5}{36} + \frac{5}{25} - 1 = \frac{5}{36} + \frac{1}{5} - 1 = \frac{25+36-180}{180} = \frac{61-180}{180} = -\frac{119}{180}$ .

Step 4: Simplify the chord equation

Setting $T=S_1$ , we get $\frac{x\sqrt{5}}{36} + \frac{y\sqrt{5}}{25} - 1 = -\frac{119}{180}$ . This simplifies to $\frac{x\sqrt{5}}{36} + \frac{y\sqrt{5}}{25} = 1 - \frac{119}{180} = \frac{180-119}{180} = \frac{61}{180}$ . Dividing by $\sqrt{5}$ , we get $\frac{x}{36} + \frac{y}{25} = \frac{61}{180\sqrt{5}}$ . This is the equation of the chord $AB$ .

Step 5: Find the length of PA and PB

Since $P$ is the midpoint of $AB$ , $PA = PB$ . Let $P=(x_1, y_1)$ . The length of the semi-chord $PA$ (or $PB$ ) can be found using the distance formula from $P$ to the intersection points $A$ and $B$ . However, a more direct approach is to use the property that for a chord passing through $P(x_1, y_1)$ , the product $PA \cdot PB = -S_1$ , where $S_1 = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1$ . Since $P$ is the midpoint, $PA = PB = \sqrt{-S_1}$ . We calculated $S_1 = -\frac{119}{180}$ . Therefore, $PA = PB = \sqrt{-\left(-\frac{119}{180}\right)} = \sqrt{\frac{119}{180}}$ .

Step 6: Calculate the final expression

We need to find $5(PA^2 + PB^2)$ . Since $PA = PB = \sqrt{\frac{119}{180}}$ , we have $PA^2 = PB^2 = \frac{119}{180}$ . Substituting these values: $5\left(\frac{119}{180} + \frac{119}{180}\right) = 5\left(2 \cdot \frac{119}{180}\right) = 5\left(\frac{119}{90}\right) = \frac{595}{90} = \frac{119}{18}$ .

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A line passing through the point P(5,5)P(\sqrt{5},\sqrt{5})P(5​,5​) intersects the ellipse x236+y225=1\dfrac{x^2}{36}+\dfrac{y^2}{25}=136x2​+25y2​=1 at points AAA and BBB such that (PA)⋅(PB)(PA)\cdot(PB)(PA)⋅(PB) is maximum. Then 5(PA2+PB2)5\bigl(PA^2+PB^2\bigr)5(PA2+PB2) is equal to: