Let the eccentricity of an ellipse [x^2 a^2 + y^2 b^2 = 1], a b, be [1 4].

Question

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 :

Accepted Answer

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

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

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

$${6 \over {{b^2}}} + {9 \over {{b^2}}} = 1 \Rightarrow {b^2} = 15$$ \& ${a^2} = 16$

${a^2} + {b^2} = 15 + 16 = 31$

Solutions · 1