JEE MAIN - Mathematics (2017 - 9th April Morning Slot - No. 14)
The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate
axes and passing through the points (4, −1) and (−2, 2) is :
$${1 \over 2}$$
$${2 \over {\sqrt 5 }}$$
$${{\sqrt 3 } \over 2}$$
$${{\sqrt 3 } \over 4}$$
Explanation
Centre at (0, 0)
$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}$$ = 1
at point (4, $$-$$ 1)
$${{16} \over {{a^2}}} + {1 \over {{b^2}}}$$ = 1
$$ \Rightarrow $$ 16b2 + a2 = a2b2 . . . .(i)
at point ($$-$$ 2, 2)
$${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$
$$ \Rightarrow $$ 4b2 + 4a2 = a2b2 . . . .(ii)
$$ \Rightarrow $$ 16b2 + a2 = 4a2 + 4b2
From equations (i) and (ii)
$$ \Rightarrow $$ 3a2 = 12b2
$$ \Rightarrow $$ a2 = 4b2
b2 = a2(1 $$-$$ e2)
$$ \Rightarrow $$ e2 = $${3 \over 4}$$
$$ \Rightarrow $$ e = $${{\sqrt 3 } \over 2}$$
$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}$$ = 1
at point (4, $$-$$ 1)
$${{16} \over {{a^2}}} + {1 \over {{b^2}}}$$ = 1
$$ \Rightarrow $$ 16b2 + a2 = a2b2 . . . .(i)
at point ($$-$$ 2, 2)
$${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$
$$ \Rightarrow $$ 4b2 + 4a2 = a2b2 . . . .(ii)
$$ \Rightarrow $$ 16b2 + a2 = 4a2 + 4b2
From equations (i) and (ii)
$$ \Rightarrow $$ 3a2 = 12b2
$$ \Rightarrow $$ a2 = 4b2
b2 = a2(1 $$-$$ e2)
$$ \Rightarrow $$ e2 = $${3 \over 4}$$
$$ \Rightarrow $$ e = $${{\sqrt 3 } \over 2}$$
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