JEE MAIN - Mathematics (2022 - 27th June Morning Shift - No. 11)
Let the eccentricity of an ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $$a > b$$, be $${1 \over 4}$$. If this ellipse passes through the point $$\left( { - 4\sqrt {{2 \over 5}} ,3} \right)$$, then $${a^2} + {b^2}$$ is equal to :
29
31
32
34
Explanation
$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
$$ \Rightarrow {{{{\left( { - 4\sqrt {{2 \over 5}} } \right)}^2}} \over {{a^2}}} + {{32} \over {{b^2}}} = 1$$
$$ \Rightarrow {{32} \over {5{a^2}}} + {9 \over {{b^2}}} = 1$$ ..... (i)
$${a^2}(1 - {e^2}) = {b^2}$$
$${a^2}\left( {1 - {1 \over {16}}} \right) = {b^2}$$
$$15{a^2} = 16{b^2} \Rightarrow {a^2} = {{16{b^2}} \over {15}}$$
From (i)
$${6 \over {{b^2}}} + {9 \over {{b^2}}} = 1 \Rightarrow {b^2} = 15$$ & $${a^2} = 16$$
$${a^2} + {b^2} = 15 + 16 = 31$$
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