Question

Find equations of both the tangent lines to the ellipse x^2+4y^2=36 that passes through the point (12,3)

Answer 1

f'( 12, 3)

Answer 2

What?

Answer 3

y' = ...?

Answer 4

So i am guessing you have to take the derivative of the ellipse to find the two equations for the tangent line and then once you have found the derivative or slope you plug in the points given to find the two equations

Answer 5

Tangent line is indeed just a the straight line: find slope by taking derivative at given point ( 12, 3)

Answer 6

The ellipse intersects x-axis at x=-6 and 6 and intersects the y-axis at y=-3 and 3. The point (12, 3) is not on the ellipse but one of the tangent lines is the horizontal line y=3 because it is tangent to the top half of the ellipse at (0, 3). Using implicit differentiation, one gets dy/dx = -x/(4y) and so letting (a, b) be the point on the ellipse for the other tangent line, which will be in the 4th quadrant, we get the following by equating the slope from the derivative to the slope of the line through (12, 3) and (a ,b):
\[\frac{ -a }{ 4b }=\frac{ 3-b }{ 12-a } \rightarrow 12b-4b^2=-12a+a^2\]
Solving the ellipse equation for y, using the negative root for the lower half of the ellipse and substituting x=a and y=b:
\[b=-\sqrt{9-x^2/4}\]
Substituting for b in the previous equation:
\[-12\sqrt{9-a^2/4}-4(9-a^2/4)=-12a+a^2\]
Simplifying to get:
\[12a-36=12\sqrt{9-a^2/4} \rightarrow (a-3)^2=9-a^2/4\]
and continuing to simplify:
\[a^2-6a=-a^2/4 \rightarrow 5a^2-24a=0 \rightarrow (5a-24)a=0\]
The solution we want is a=24/5 which yields b=-9/5 and a slope for the line of 2/3 and so the equation of the second tangent line is:
\[y-3=\frac{ 2 }{ 3 }(x-12) \rightarrow y=\frac{ 2 }{ 3 }x-5\]