Let the foci of the Ellipse x 2 16 + y 2 7 = 1 and Hyperbola x 2 144 - y 2 = 1 25 coincide. Then the length of the latus rectum of the Hyperbola is:

Question

Let the foci of the Ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and Hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the Hyperbola is:

Smart Achievers · Accepted Answer

(2)

$Foci of the Ellipse = (\pm 3, 0)$

$Foci of the Hyperbola = \pm \frac{1}{5} \sqrt{144 + α}$ $∴$ $α = 81$

$length of the latus rectum = \frac{2 b^{2}}{a} = \frac{27}{10}$

Solution

Q.
Let the foci of the Ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and Hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the Hyperbola is:

1 $\frac{32}{9}$

2 $\frac{27}{10}$

3 $\frac{18}{5}$

4 $\frac{27}{4}$

Ans.
(2)

$Foci of the Ellipse = (\pm 3, 0)$

$Foci of the Hyperbola = \pm \frac{1}{5} \sqrt{144 + α}$ $∴$ $α = 81$

$length of the latus rectum = \frac{2 b^{2}}{a} = \frac{27}{10}$

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Solution

Q. Let the foci of the Ellipse x216+y27=1 and Hyperbola x2144-y2α=125 coincide. Then the length of the latus rectum of the Hyperbola is:

1 329

2 2710

3 185

4 274

Ans. (2) Foci of the Ellipse=(±3,0) Foci of the Hyperbola=±15144+α ∴ α=81 length of the latus rectum=2b2a=2710