JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 11)
Let S = $$\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.$$ Then S represents :
an ellipse whose eccentricity is $${1 \over {\sqrt {r + 1} }},$$ where r > 1
an ellipse whose eccentricity is $${2 \over {\sqrt {r + 1} }},$$ where 0 < r < 1
an ellipse whose eccentricity is $${2 \over {\sqrt {r - 1} }},$$ where 0 < r < 1
an ellipse whose eccentricity is $$\sqrt {{2 \over {r + 1}}}$$, where r > 1
Explanation
$${{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}} = 1$$
for r > 1, $${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$$
$$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)} $$
$$ = \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}} $$
$$ = \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}} $$
for r > 1, $${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$$
$$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)} $$
$$ = \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}} $$
$$ = \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}} $$
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