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Hyperbola question

2020 · 3 Sep · Shift 2 · Q27
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  5. /2020 · 3 Sep · Shift 2 · Q27

Hyperbola question

2020 · 3 Sep · Shift 2 · Q27

JEE MainMathematicsHyperbolaMCQ+4 / −1
Let e1 and e2 be the eccentricities of the ellipse, x225+y2b2=1{{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 125x2​+b2y2​=1(b < 5) and the hyperbola, x216−y2b2=1{{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 116x2​−b2y2​=1 respectively satisfying e1e2 = 1. If α\alphaα and β\betaβ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (α\alphaα, β\betaβ) is equal to :
  1. A
    (8, 10)
  2. B
    (8, 12)
  3. C
    (245,10)\left( {{{24} \over 5},10} \right)(524​,10)
  4. D
    (203,12)\left( {{{20} \over 3},12} \right)(320​,12)
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