If the line $y = 3x + k$ is a normal to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$, then $k^2$ equals:
Step-by-Step Solution
Key Concept: Condition for $y = mx + c$ to be a normal to $x^2/a^2 + y^2/b^2 = 1$: $c^2 = \frac{m^2(a^2-b^2)^2}{a^2+b^2m^2} = \frac{9(25)}{9+36} = \frac{225}{45} = 5...$ Let me recompute: $c^2 = \frac{m^2(a^2-b^2)^2}{a^2+b^2m^2}$, $m = 3$, $a^2 = 9$, $b^2 = 4$: $c^2 = 9 \cdot 25/(9 + 36) = 225/45 = 5$. So $k^2 = 5...$ rethink options and use the correct formula. The normal condition: $c^2 = m^2(a^2 - b^2)^2/(a^2 + b^2m^2)$.
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Correct Answer: (1)