What is an equation of a line tangent to 2x^2 + y^2 -4x -3y -16 = 0 that contains the point (-3.5, 6)?

Question

What is an equation of a line tangent to 2x²+y²-4x-3y-16=0 that contains the point (-3.5, 6)?

The point is in the tangent line, not the equation of the oval.

I found the derivative of the oval to be (4x-4)/(3-2y) if that helps.

Answer 1

There is actually a 2nd tangent line passing through (-3.5,6) that is tangent to the ellipse. Here would be a general strategy for finding the equations of these tangent lines (without relying strictly on graphing. It is likely less confusing to use the letters (a,b) to represent the point on the ellipse that has a tangent line passing through the target point.

Then the slope of the tangent line at (a,b) passing through (-3.5,6) is (b-6)/a+3.5). But also using the formula you discovered for y' the slope at (a.b) must be (4a-4)/(3-2b). Set those expressions equal to each other, cross multiply to get something like: 4a²+2b²+10a-15b-4 = 0. Since (a,b) is on the ellipse, this equation is also true:

2a²+b²-4a-3b-16 = 0. So we have a system of equations that can be solved to find values for a and b. If you are not sure how to solve such a system, do a search for "finding points of intersection for two ellipses (or circles for that matter). In the end you will find a = 1 or a=-2. So points on the ellipse that have a tangent line passing through (-3.5,6) are (1,6) and (-2,0). You can calculate the slope between (-3.5,6) and (-2,0) and then use point-slope to get the 2nd tangent line equation.

Answer 2

Graph the ellipse.

One end of the major axis is (1, 6)

The equation of the tangent line is y = 6