Which of the following is the slope form of a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$?
A$y=mx\pm\sqrt{a^2m^2+b^2}$
B$y=mx\pm\sqrt{a^2+b^2m^2}$
C$y=mx\pm\sqrt{a^2m^2-b^2}$
D$y=mx\pm\sqrt{b^2m^2-a^2}$
Answer & Solution
Correct answer: A. $y=mx\pm\sqrt{a^2m^2+b^2}$
Take the line $y=mx+c$. For tangency with the ellipse, substituting into $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ must give a quadratic in $x$ with discriminant zero. This leads to $c^2=a^2m^2+b^2$, so $c=\pm\sqrt{a^2m^2+b^2}$. Hence the tangent in slope form is $y=mx\pm\sqrt{a^2m^2+b^2}$.
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