The ellipse $$\(E\)$$ has equation $$\[ \frac{x^{2}}{49}+\frac{y^{2}}{b^{2}}=1 \]$$ where $$\(b\)$$ is a constant and $$\(0< b< 7\)$$ The eccentricity of the ellipse is $$\(e\)$$ Hence show that $$\[ b^{2}=49\left(1-e^{2}\right) \]$$ (2)

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