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A point $P(x,y)$ is tested against the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Which condition correctly identifies that $P$ lies outside the ellipse?
A$\frac{x^2}{a^2}+\frac{y^2}{b^2}<1$
B$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
C$\frac{x^2}{a^2}+\frac{y^2}{b^2}>1$
D$\frac{x^2}{a^2}-\frac{y^2}{b^2}>1$
Answer & Solution
Correct answer: C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}>1$
For the standard ellipse, points on the curve satisfy $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If the left-hand side is less than 1, the point is inside; if it equals 1, the point is on the ellipse; and if it is greater than 1, the point is outside. Therefore option C is correct.
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