Let e1 and e2 be the eccentricities of the ellipse, [x^2 25 + y^2 b^2 = 1(b

Question

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 :

Accepted Answer

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)

Solutions · 1