Answer to Question #273306 in Civil and Environmental Engineering for deshi

Question

Find the tangent lines as directed to the parabola (x+2y)^2+2x-y-3=0 parallel to the line 4x+3y=2

Answer 1

Let the tangent be x-4y=k => x= k+4y. Use this in the equation of the curve

$(4y+k)^2 +2(k+4y) -y -3 = 16y^2 +(8k+7)y +(k^2+2k-3)=0$

This should have equal roots so 

$(8k+7)^2–4(16)(k^2+2k-3)=9 or (112–228)k+(49+192)=0$

k= 241/116