The equation of the straight lines joining the foci of the ellipse x^2/25 + y^2/16 = 1 to the foci of the ellipse ​

Question

The equation of the straight lines joining the foci of the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) to the foci of the ellipse \(\frac{x^2}{24} + \frac{y^2}{49} = 1\) forms a parallelogram. Then the area of the figure formed by the foci of these two ellipse.

(a) 15

(b) 30

(c) 20

(d) 18

Comments

Need solution

Answer 1

Correct option is (b) 30

Answer 2

Correct Option is (b) 30

Answer 3

Correct option is (b) 30

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

a = 5 

b = 4

a > b

\(b^2 = a^2 (1-e^2)\)

\(16 = 25 (1-e^2)\)

e = 3/5

focus \((\pm ae ,0) \equiv (\pm 3,0)\)

\(\frac {x^2}{24} + \frac {y ^2}{49} =1\)

a \(=\sqrt {24}\)

b = 7

a < b

\(a^2 = b^2 (1-e ^2)\)

\(24 = 49 (1-e^2)\)

e = 5/7

focus \((0, \pm be) \equiv (0, \pm 5)\)

Let diagonals be d₁ and d₂

\(d_1 = \sqrt {(0-0)^2 + (5+5)^2} =10\)

\(d_2 = \sqrt {(-3-3)^2 + (0-0)^2 }= 6\)

Now, Area of parallelogram

\(=\frac {1}{2}\times d_1 \times d_2\)

\( =\frac {1}{2} \times 10 \times 6\)

= 30