Diff Cal thanks for the help

Question

find the tangent and normal to the ellipse x^2 + 9y^2-2x-6y+2=0, at (1, 1/3). y"=___, equation of tangent is ____, equation of normal line is____

Answer 1

x^2 + 9y^2 -2x -6y + 2 = 0

x^2 -2x + 1 + 9(y^2 - 6y/9 + 1/9) = -2 +1 +1 = 0

(x-1)^2 + 9(y-1/3)^2 = 0

(x-1)^2/1^2 + (y-1/3)^2/(1/3)^2 = 0

this is a degenerate ellipse, degenerating to a single point at (1, 1/3), its center. Any "tangent" line through that point would have an indeterminate slope, or any slope, such as

1 or 0. for slope =0, the "tangent line" is y=1/3, for slope = 1, the "tangent line" would be y=x- 2/3. Or for an undefined slope, the "tangent line" would be x=1.

if the right side constant term had been 1, instead of 0, then it would be an actual ellipse, with major semi-axis = 1 with endpoints (-2,1/3) and (0, 1/3), and major semi-axis = 1/3 with vertical endpoints (1,-2/3) and (1, 0)

if the constant term had been 1 instead of 2

it would have been an actual ellipse

still (1, 1/3) is the ellipse's center and not a point on the ellipse, so there is no tangent line at that point.

tangent lines are easily found at the vertical and horizontal endpoints. y=0, y=-2/3, x= 0 and x=-2 for the top, bottom ellipse points and the right and left end points. (1,0),(1,-2/3), (0,1/3) and (-2,1/3). slopes are zero, zero, undefined and undefined.

At any rate, a tiny change in the problem, here or there, and it's a more solvable problem. To visually see the "degenerate ellipse" or degenerate circle, try graphing x^2+y^2 = 1, then watch the circle slowly collapse into a point as you graph x^2+y^2 = 0.01, then = 0.001. It degenrates into a single point at the center. similar to watching an obtuse triangle collapse into a straight line as the obtuse angle approaches 180 degrees.