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Given equation $x^2+9y^2=81$ and the point $(27,3)$, find the equation of 2 lines that pass through the point $(27,3)$, and is tangent to the ellipse

so by using implicit differentiation I got $y'=\frac{-x}{9y}$, which is the slope of the line. but i don't know where to go from here. Any help would be appreciated.

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Let $f: (x,y) \mapsto x^{2}+9y^{2}: \mathbb{R}^{2} \to \mathbb{R}$; then $f^{(-1)}\{ 81 \}$ is the ellipse under consideration. Note that $\nabla f(27,3) = (54,54)$ is normal to $f^{(-1)}\{ 81 \}$, so $(x,y)$ lies on the tangent line through $(27,3)$ iff $$ (x-27,y-3)\cdot (54,54) = 0, $$ i.e. iff $$ x+y = 30. $$

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Do you mean find the equation of the tangent line at $(27,3)$? You just use point slope formula $y-3=\frac{dy}{dx}(x-27)$.

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