## Calculus 8th Edition

$y=3$ $y=\frac{2}{3}x-5$
$x^2+4y^2=36...(a)\\ 2x+8yy'=0\\ y'=-\frac{x}{4y}$ Equation of a tangent line at $(12,3)$ $y-3=m(x-12)\\ y-3=-\frac{x}{4y}(x-12)\\ y-3=\frac{-x^2+12x}{4y}\\ 4y^2-12y=-x^2+12x\\ 4y^2+x^2=12y+12x$ From $(a)$ $36=12y+12x\\ 3=x+y\\ y=3-x...(b)$ Plug in $(b)$ into $(a)$ $x^2+4(3-x)^2=36\\ x^2+4(9+x^2-6x)=36\\ x^2+36+4x^2-24x=36\\ 5x^2-24x=0\\ x(5x-24)=0\\ x=0\Longrightarrow y=3\\ x=\frac{24}{5}\Longrightarrow y=-\frac{9}{5}$ First tangent line $\frac{y-3}{3-3}=\frac{x-0}{12-0}\\ y-3=0\\ y=3$ Second tangent line $\frac{y-(-\frac{9}{5})}{3-(-\frac{9}{5})}=\frac{x-\frac{24}{5}}{12-\frac{24}{5}}\\ y=\frac{2}{3}x-5$