Let P(x, y) be any point on the ellipse 25 x^2+16 y^2=400 . If Q(0,3) and R(0,-3) are two points, then what is (P Q+P R) equal to?

Step-by-step solution

Equation of the Ellipse : 25 x^2+16 y^2=400 . x^2 16 + y^2 25 =1 Here a^2=16, b^2=25 aligned Focci points & =(0, b e)= (0, b^2-a^2 ) & =(0, 25-16 )=(0, 3) aligned Point Q=(0,3) , and R=(0,-3) . Now, from question point P(x, y) lies on the ellipse, then sum of its distance from two fixed points is always constant, equal to length of its major axis. Thus, (P Q+P R)=2 5=10 .