A line passing through the point P(5,5) intersects the ellipse 36x2+25y2=1 at points A and B such that (PA)⋅(PB) is maximum. Then 5(PA2+PB2) is equal to:
Get the complete, step-by-step math solution for: "A line passing through the point P(√(5),√(5)) intersects the ellipse {x²}{36}+ {y²}{25}=1 at points A and B such that (PA)·(PB) is maximum. Then 5 (PA...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Understand the problem and the power of a point theorem
The problem asks us to find 5(PA2+PB2) for a line passing through point P(5,5) and intersecting the ellipse 36x2+25y2=1 at points A and B. The condition is that the product PA⋅PB is maximum. For a point P outside an ellipse, the product PA⋅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(5,5) is inside the ellipse, as shown by substituting its coordinates into the ellipse equation: 36(5)2+25(5)2=365+255=36/51+51=71.2+51=0.138+0.2=0.338<1. Therefore, the line segment AB is a chord passing through P. The product PA⋅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 (5,5). The chord AB for which P is the midpoint is given by the equation T=S1, where T is the tangent equation and S1 is the value of the ellipse equation at point P. The equation of the ellipse is 36x2+25y2−1=0.
Step 3: Calculate T and S1
Let's calculate T and S1. The general form for T for an ellipse a2x2+b2y2=1 at a point (x1,y1) is a2xx1+b2yy1=1. So, for P(5,5), T is 36x5+25y5−1. The value S1 is 36(5)2+25(5)2−1=365+255−1=365+51−1=18025+36−180=18061−180=−180119.
Step 4: Simplify the chord equation
Setting T=S1, we get 36x5+25y5−1=−180119. This simplifies to 36x5+25y5=1−180119=180180−119=18061. Dividing by 5, we get 36x+25y=180561. 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=(x1,y1). 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(x1,y1), the product PA⋅PB=−S1, where S1=a2x12+b2y12−1. Since P is the midpoint, PA=PB=−S1. We calculated S1=−180119. Therefore, PA=PB=−(−180119)=180119.
Step 6: Calculate the final expression
We need to find 5(PA2+PB2). Since PA=PB=180119, we have PA2=PB2=180119. Substituting these values: 5(180119+180119)=5(2⋅180119)=5(90119)=90595=18119.