Question
If one end of a diameter of the ellipse $4 x^{2}+y^{2}=64$ is at $(2 \sqrt{3}, 4)$, then the other end is at(a) $(-2 \sqrt{3},-4)$(b) $(2 \sqrt{3},-4)$(c) $(-2 \sqrt{3}, 4)$(d) $(2 \sqrt{3}, 4)$
Step 1
The standard form of an ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Comparing these two equations, we can see that $a^{2}=16$ and $b^{2}=64$. Therefore, the semi-major axis is $a=4$ and the semi-minor axis is $b=8$. Show more…
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