Asked by Anonymous
Find the equations for the lines tangent to the ellipse 4x^2+y^2=72 that are perpendicular to the line x+2y+3=0
Answers
Answered by
drwls
If a line is perpendicular to
x+2y+3=0 (or y = -x/2 - 3/2),
the slope of the line must be 2.
Solve the equation dy/dx = 2.
y = sqrt(72 - 4x^2)
dy/dx = (-4x)/sqrt(72 - 4x^2) = 2
x+2y+3=0 (or y = -x/2 - 3/2),
the slope of the line must be 2.
Solve the equation dy/dx = 2.
y = sqrt(72 - 4x^2)
dy/dx = (-4x)/sqrt(72 - 4x^2) = 2