JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 6)
Let e1
and e2
be the eccentricities of the
ellipse,
$${{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1$$(b < 5) and the hyperbola,
$${{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1$$ respectively satisfying e1e2 = 1. If $$\alpha $$
and $$\beta $$ are the distances between the foci of the
ellipse and the foci of the hyperbola
respectively, then the ordered pair ($$\alpha $$, $$\beta $$) is equal to :
$${{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1$$(b < 5) and the hyperbola,
$${{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1$$ respectively satisfying e1e2 = 1. If $$\alpha $$
and $$\beta $$ are the distances between the foci of the
ellipse and the foci of the hyperbola
respectively, then the ordered pair ($$\alpha $$, $$\beta $$) is equal to :
(8, 10)
(8, 12)
$$\left( {{{24} \over 5},10} \right)$$
$$\left( {{{20} \over 3},12} \right)$$
Explanation
For ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1\,\,\,(b < 5)$$
Let e1 is eccentricity of ellipse
$$ \therefore $$ b2 = 25 (1 $$ - $$ $${e_1}^2$$) ........(1)
Again for hyperbola
$${{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1$$
Let e2 is eccentricity of hyperbola.
$$ \therefore $$ $${b^2} = 16({e_1}^2 - 1)$$ ......(2)
by (1) & (2)
$$25(1 - {e_1}^2)\, = \,16({e_1}^2 - 1)$$
Now e1 . e2 = 1 (given)
$$ \therefore $$ $$25(1 - {e_1}^2)\, = \,16\left( {{{1 - {e_1}^2} \over {{e_1}^2}}} \right)$$
or e1 = $${4 \over 5}$$ $$ \therefore $$ e2 = $${5 \over 4}$$
Now distance between foci is 2ae
$$ \therefore $$ Distance for ellipse = $$2 \times 5 \times {4 \over 5} = 8 = \alpha $$
Distance for hyperbola = $$2 \times 4 \times {5 \over 4} = 10 = \beta $$
$$ \therefore $$ ($$\alpha $$, $$\beta $$) $$ \equiv $$ (8, 10)
Let e1 is eccentricity of ellipse
$$ \therefore $$ b2 = 25 (1 $$ - $$ $${e_1}^2$$) ........(1)
Again for hyperbola
$${{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1$$
Let e2 is eccentricity of hyperbola.
$$ \therefore $$ $${b^2} = 16({e_1}^2 - 1)$$ ......(2)
by (1) & (2)
$$25(1 - {e_1}^2)\, = \,16({e_1}^2 - 1)$$
Now e1 . e2 = 1 (given)
$$ \therefore $$ $$25(1 - {e_1}^2)\, = \,16\left( {{{1 - {e_1}^2} \over {{e_1}^2}}} \right)$$
or e1 = $${4 \over 5}$$ $$ \therefore $$ e2 = $${5 \over 4}$$
Now distance between foci is 2ae
$$ \therefore $$ Distance for ellipse = $$2 \times 5 \times {4 \over 5} = 8 = \alpha $$
Distance for hyperbola = $$2 \times 4 \times {5 \over 4} = 10 = \beta $$
$$ \therefore $$ ($$\alpha $$, $$\beta $$) $$ \equiv $$ (8, 10)
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