Question
The length of the latus rectum of the hyperbola whose equation is $x^{2}-4 y^{2}$ $=16$ is(A) 1(B) 2(C) $\sqrt{20}$(D) 2$\sqrt{20}$(E) 16
Step 1
We can rewrite this equation in the standard form of a hyperbola by dividing both sides by 16. This gives us $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$. Show more…
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