The equation of the directrices of the ellipse 16 x 2-25 y 2-400 areA- 3 x- 10B- 5 x- 9C- 3 x- 25D- 2 x- 25

The correct option is C x^2/25 + y^2/16 = 1 ⇒ e = √(1 16/25) = 3/5 nbsp; Therefore , directrices are x nbsp;±5/3/5 = 0 or 3x nbsp;± 25 = 0 . nbsp; ...

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Byju's Answer

Standard XI

Mathematics

Intercept of a Straight Line

The equation ...

Question

The equation of the directrices of the ellipse $16 x^{2} + 25 y^{2} = 400$ are

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Solution

The correct option is D

$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 \Rightarrow e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$

Therefore , directrices are x ± $\frac{5}{\frac{3}{5}} = 0 o r 3 x \pm 25 = 0 .$

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Similar questions

Q. The line

3 x + 5 y = k

is a tangent to the ellipse

16 x^{2} + 25 y^{2} = 400,

k

is:

Q. Which of the following statements is correct?

I .

Equation of the director circle with respect to the ellipse

16 x^{2} + 25 y^{2} = 400

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

I I .

Equation of the tangent at

(3, 2)

on the ellipse

x^{2} + 4 y^{2} = 25

3 x + 8 y - 25 = 0

Q. A ray eminating from the point

(- 3, 0)

is incident on the ellipse

16 x^{2} + 25 y^{2} = 400

at the point

P

with ordinate

4 .

Equation of the reflected ray is

Q. If PSQ is a focal chord of the ellipse

16 x^{2} + 25 y^{2} = 400

such that SP = 8 then the length of SQ =

Q. Find by integration, the are of the ellipse

16 x^{2} + 25 y^{2} = 400

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The equation of the directrices of the ellipse 16x2+25y2=400 are

The correct option is D x225+y216=1⇒e=√1−1625=35 Therefore , directrices are x ±535=0 or 3x±25=0.

The equation of the directrices of the ellipse $16 x^{2} + 25 y^{2} = 400$ are

The correct option is D

$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 \Rightarrow e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$

Therefore , directrices are x ± $\frac{5}{\frac{3}{5}} = 0 o r 3 x \pm 25 = 0 .$