Question 1183554
{{{x^2/16 + y^2/9 = 1}}}  ===> {{{x/16 + (y*(dy/dx))/9=0}}} ===> {{{dy/dx = -(9x)/(16y)}}}.

Now let ({{{x[0]}}}, {{{y[0]}}}) be the point of tangency in the 1st quadrant.  The tangent line should be 

{{{y - y[0] = -((9x[0])/(16y[0]))*(x-x[0])}}}.

The y-intercept of this is {{{y = 9/y[0]}}}.  (Verify!)

And the x-intercept is {{{x = 16/x[0]}}}. (Verify!)

The area of the triangle in the 1st quadrant is then {{{A = 72/(x[0]*y[0]) = 96/(x[0]*sqrt(16 - x[0]^2))}}}.

Now get the derivative of A wrt {{{x[0]}}}, and set it to 0:

{{{dA/dx[0] = -96*( (sqrt(16-x[0]^2) - x[0]^2/sqrt(16-x[0]^2) ) / (x[0]^2*(16 - x[0]^2)) ) = -192*(8 - x[0]^2)/(x[0]^2*(16 - x[0]^2)^(3/2)) = 0}}}

===> {{{x[0] = 2sqrt(2)}}}, so there is a local extremum at this point.

If {{{x < 2sqrt(2)}}}, then {{{dA/dx[0] < 0}}}. (E.g., choose {{{x=sqrt(2)}}}.)

If {{{x > 2sqrt(2)}}}, then {{{dA/dx[0] > 0}}}. (E.g., choose {{{x=3}}}.)

Hence there is absolute minimum for the area in the 1st quadrant at {{{x[0] = 2sqrt(2)}}}, by the 1st derivative test.
The corresponding y-value is {{{y[0] = (3sqrt(2))/2}}}.

The equation of the tangent line is then {{{y - y[0] = -((9x[0])/(16y[0]))*(x-x[0])}}}, or {{{highlight(y - (3sqrt(2))/2 = -(3/4)*(x-2sqrt(2)))}}}.


The area of the triangle of least area is {{{A = 72/(x[0]*y[0]) = highlight(12)}}}