Calculus problem? Find all tangent lines to the ellipse x^2+4*y^2=5 which pass through the point (-5,0).?

Question

The solution should describe how you know you have found all of the solutions.

Help please?

Answer 1

y= m(x+5) I interchanged coordinates formerly
x^2+4m^2(x+5)^2-5=0
x^2(1+4m^2) +40m^2x +100m^2-5=0
Double root
1600m^4 -4(1+4m^2)(100m^2-5)=0
320m^2-20=0 so m=+-1/4
y=+-1/4(x+5)
Through a point there are at most two tangents

Answer 2

I think of the sole ingredient you alter is the sign of the slope... yet i need to be incorrect. Take spinoff implicitly 2x + 8y(dy/dx) = 0 resolve for y as a function of x 2x = -8y(dy/dx) -2x/8y = dy/dx plug contained in the criteria (-2(12))/(8*3) = -one million . what a coinkydink so dy/dx = -one million at those factors; slope of the tangent line. making use of equation of tangent line formulation; y - 3 = -one million(x - 12) uh. swap the sign of the slope in case you like yet another line. in any different case, it rather is it.. ((must be incorrect))