Question

8.
Let P(x, y) be any point on the ellipse
25x2 + 16y2 = 400. If Q(0, 3) and R(0, -3) are
two points, then what is (PQ + PR) equal to ?
(a)
12
(b)
10
(c)
8
(d)
6​

Answer 1

It has given that, P(x, y) be any point on the ellipse 25x² + 16y² = 400. here Q(0, -3) and R(0, -3) are two points.

To find : what is the value of PQ + PR equals to..

solution : equation of ellipse, 25x² + 16y² = 400

⇒25x²/400 + 16y²/400 = 1

⇒x²/16 + y²/25 = 1

⇒x²/(4)² + y²/(5)² = 1

on comparing x²/a² + y²/b² = 1

a = 4, b = 5

here a < b so the major axis of ellipse is along y - axis and length of major axis = 2b = 2 × 5 = 10

foci = (0, ± c)

c = √(b² - a²) = √(5² - 4²) = 3

so, foci = (0, ±3)

hence P(0, 3) and R(0, -3) are foci of given ellipse.

we know sum of distance of any point on ellipse to foci always remains same and equals to the length of major axis.

i.e., PQ + PR = major axis = 10

Therefore the value of PQ + PR is 10 unit. so the option (b) is correct choice.