the line has slope 2/3
So, the normal needs slope -3/2
3x^2-4y^2=8
6x - 8yy' = 0
y' = 3x/4y
So, we need 3x/4y = -3/2
y = -x/2
3x^2-4(-x/2)^2 = 8
3x^2-x^2 = 8
x = ±2
y = ∓1
So, we want lines through (2,-1) and (-2,1) with slope -3/2
y-1 = -3/2 (x+2)
y+1 = -3/2 (x-2)
See the graphs at
http://www.wolframalpha.com/input/?i=plot+3x^2-4y^2%3D8%2C+2x-3y%2B6%3D0%2C+y+%3D+-3%2F2+%28x%2B2%29%2B1%2Cy+%3D+-3%2F2+%28x-2%29-1
Find the equation of the tangent line to the hyperbola 3x^2-4y^2=8 which is/are perpendicular to the line 2x-3y+6=0. Give the answers in general form.
1 answer
1 answer