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If the length...

Question

If the length of perpendicular drawn from origin to any normal to the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$ is $l$ , then $l$ cannot be

4

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5 / 2

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1 / 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 / 3

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Solution

The correct options are
A $4$
B $5 / 2$
C $1 / 2$
D $2 / 3$

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\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1

a + 5

a

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

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l =

x^{2} + y^{2} = 41

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

x^{2} + y^{2} = 25

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1,

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

l

l

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