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UPSC NDA I 2026 Mathematics Paper

Question Numbers: 61-62

For the next two (02) items that follow:

The foci of the ellipse

\mathrm{px}^{2}+16 \mathrm{y}^{2}=16 \mathrm{p}

and the foci of the hyperbola

25\left(81 x^{2}-144 y^{2}\right)=11664

coincide (assume

\mathrm{p} < 16

Question : 62 of 120

Marks: +1, -0

Solution:

Concept:

Since the foci of the ellipse and hyperbola coincide, their focal distances from the centre are equal.

Explanation:

Ellipse:

px^{2}+16y^{2}=16p

Dividing by

16p

\frac{x^{2}}{16}+\frac{y^{2}}{p}=1

a=4

b^{2}=p

, hence

c_e^{2}=16-p

Hyperbola:

25(81x^{2}-144y^{2})=11664

81x^{2}-144y^{2}=\frac{11664}{25}

\frac{x^{2}}{144/25}-\frac{y^{2}}{81/25}=1

a_h=\frac{12}{5}

b_h=\frac{9}{5}

, hence

c_h^{2}=\frac{144+81}{25}=9

, giving

c_h=3

Since foci coincide,

c_e=3

\sqrt{16-p}=3 \Rightarrow p=7

Ellipse eccentricity:

e_e=\frac{3}{4}=0.75

Hyperbola eccentricity:

e_h=\frac{3}{12/5}=\frac{5}{4}=1.25

Required difference:

1.25-0.75=0.5

Answer:

0.5

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