What is the length of the latus rectum of the ellipse $25 x^{2} + 16 y^{2} = 400 ?$

\frac{25}{2}

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\frac{25}{4}

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\frac{16}{5}

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\frac{32}{5}

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Solution

The correct option is C $\frac{25}{2}$
$\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$

hence,
$a = 4, b = 5$

length of latus rectum = $2 \frac{b^{2}}{a} = \frac{25}{2}$

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Q. Find the equation of axes, driectrix, co-ordinate of focii, centre, vertices, length of latus - rectum and eccentricity of an ellipse

\frac{(x - 3)^{2}}{25} + \frac{(y - 2)^{2}}{16} = 1

Q. The eccentricity of the ellipse 25x² + 16y² = 400 is
(a) 3/5
(b) 1/3
(c) 2/5
(d) 1/5

Q. The eccentricity of the ellipse

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

Q. The following equation represents an ellipse,

25 (x^{2} - 6 x + 9) + 16 y^{2} = 400

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

Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i) $4 x^{2} + 9 y^{2} = 1$
(ii) $5 x^{2} + 4 y^{2} = 1$
(iii) $4 x^{2} + 3 y^{2} = 1$
(iv) $25 x^{2} + 16 y^{2} = 1600$
(v) $9 x^{2} + 25 y^{2} = 225$

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What is the length of the latus rectum of the ellipse 25x2+16y2=400?

The correct option is C 252x216+y225=1hence,a=4,b=5length of latus rectum =2b2a=252

What is the length of the latus rectum of the ellipse $25 x^{2} + 16 y^{2} = 400 ?$

The correct option is C $\frac{25}{2}$
$\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$

hence,
$a = 4, b = 5$

length of latus rectum = $2 \frac{b^{2}}{a} = \frac{25}{2}$