The foci of the ellipse 4 x^2+9 y^2=1 are at Q and R. If P(x, y) is any point on the ellipse, then what is PQ+PR equal to?

Correct answer: B. 1 Given equation of ellipse, 4 x^2+9 y^2=1 Here, a=(1)/(2) and b=(1)/(3) Since, P(x, y) is any point on the ellipse. So, PQ+PR=2 a=2 × (1)/(2)=1 .

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Ellipse

NDA 2024 · Mathematics · Ellipse

The foci of the ellipse 4 x^2+9 y^2=1 are at Q and R. If P(x, y) is...

The foci of the ellipse

4 x^2+9 y^2=1

are at Q and R. If

\mathrm{P}(x, y)

is any point on the ellipse, then what is

\mathrm{PQ}+\mathrm{PR}

equal to?

Options

A $2$
B $1$
C $\frac{2}{3}$
D $\frac{1}{3}$

Correct answer

B. $1$

Step-by-step solution

Given equation of ellipse,

4 x^2+9 y^2=1

Here,

\mathrm{a}=\frac{1}{2}

and

\mathrm{b}=\frac{1}{3}

Since,

\mathrm{P}(\mathrm{x}, \mathrm{y})

is any point on the ellipse. So,

\mathrm{PQ}+\mathrm{PR}=2 \mathrm{a}=2 \times \frac{1}{2}=1

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