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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 : 61 of 120

Marks: +1, -0

What is the value of

p

Solution:

Concept:

For ellipse and hyperbola, foci distances from centre are

\sqrt{a^2-b^2}

and

\sqrt{a^2+b^2}

respectively; equate them.

Explanation:

Ellipse:

px^2+16y^2=16p

Divide by

16p

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

Since

p<16

a^2=16

b^2=p

. Foci are

(\pm\sqrt{16-p},0)

Hyperbola:

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

Rewrite as

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

Thus

a^2=\frac{144}{25}

b^2=\frac{81}{25}

Focus distance:

c=\sqrt{a^2+b^2}=\sqrt{\frac{225}{25}}=3

So hyperbola foci are

(\pm3,0)

Equating foci:

\sqrt{16-p}=3

16-p=9\Rightarrow p=7

Answer:

p=7

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