Question

find the tangent line to the equation x3+y3=6xy at the point (3,3) and at what point the tangent line horizontal in the first quadrant​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given the equation of a curve, find the equation of the tangent at point $(3,3)$

Explanation:

1. given an equation of the curve $y=f(x)$ and a point $(x',y')$ that lies on the curve we can determine the slope of the tangent at the given on the curve by,     $m=\frac{dy}{dx}|_{(x=x',y=y')}$         -----(a)                                                                        
2. from above point we can find the equation of the tangent line using ,       $y-y'=m(x-x')$     -----(b)        
3. here we have,  $x^3+y^3=6xy$  
4. taking implicit derivative we get,                                                                             [tex]->3x^2dx+3y^2dy=6xdy+6ydx\\ ->(x^2-2y)dx=(2x-y^2)dy\\ ->\frac{dy}{dx} =\frac{x^2-2y}{2x-y^2} \\ ->\frac{dy}{dx}|_{(x=3,y=3)} =\frac{3^2-2(3)}{2(3)-3^2} \\ ->\frac{dy}{dx}|_{(x=3,y=3)} =-1 \\[/tex]    
5. putting this in (b) we get the equation of tangent at $(3,3)$,                                            [tex]->y-3=-1(x-3)\\ ->y+x-6=0[/tex]           ------->ANSWER
6. for the tangent to be horizontal the slope should be zero , hence                  [tex]->\frac{dy}{dx}=0\\ ->\frac{x^2-2y}{2x-y^2}=0 \\ ->y=\frac{x^2}{2} [/tex]  
7. equating this in the equation of the curve we get,                                             [tex]->x^3+\frac{x^6}{8}=3x^3\\ ->x^6-16x^3=0\\ ->x^3(x^3-16)\\ ->x=0, 2\sqrt[3]{2} [/tex]    
8. hence, putting these values back in the curve equation we get,                        [tex]x= 0\ \ \ \ \ \ \ ->y=0\\ x=2\sqrt[3]{2}\ \ \ -> y=2\sqrt[3]{4} [/tex]  
9. the points in first quadrant at which the tangent is horizontal are $(0,0)\ and\ (2\sqrt[3]{2}, 2\sqrt[3]{4} )$