"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 :"

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Accepted Answer

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${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$

$$ \Rightarrow {{{{\left( { - 4\sqrt {{2 \over 5}} } \right)}^2}} \over {{a^2}}} + {{32} \over {{b^2}}} = 1$$

$\Rightarrow {{32} \over {5{a^2}}} + {9 \over {{b^2}}} = 1$ ..... (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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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 :

Solution

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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 :

Solution

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