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UPSC NDA I 2023 Mathematics Solved Paper

Question Numbers: 57-58

Consider the following for the next two (02) items that follow :

P(x, y) is any point on the ellipse

x^{2}+4y^{2}=1

. Let E, F be the foci of the ellipse.

Question : 57 of 120

Marks: +1, -0

What is PE + PF equal to ?

Solution:

Concept:

For an ellipse, the sum of distances from any point on the ellipse to its two foci is constant and equals

2a

, where

a

is the length of the semi‑major axis.

Explanation:

The given ellipse is

x^2 + 4y^2 = 1

Rewrite in standard form:

\frac{x^2}{1} + \frac{y^2}{\frac14} = 1

Hence

a^2 = 1

and

b^2 = \frac14

a = 1

and

b = \frac12

Let

E

and

F

be the foci.

For any point

P(x,y)

on the ellipse, the ellipse property gives

PE + PF = 2a

Therefore,

PE + PF = 2(1) = 2

Answer:

2 (Option B).

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