Thread: Algebraic proof

An ellipse has the equation x^2+5y^2=5
a line has the equation y=mx+c

(1) Show that if the line is a tangent to the ellipse then c^2=5m^2+1
(2) hence find the equation of the tangent parallel to the line x-2y+1=0

(note ^ symbol: is the to the power of)

If anyone could help, I would much appreciate it. Thankyou

Hey Mathshelp246.

Hint: First try finding dy/dx through implicit differentiation (this corresponds to your m value).

okay for the problem I got dy/dx to be 2x+10y. dy/dx=0 therefore dy/dx=-2x/10y which simplifies to dy/dx= -x/5y so m= -x/5y

found c to be : y=mx+c sub in the m value to be y=(-x/5y)x+c therefore y=-x^2/5y+c so, c=y+(x^2/5y) = 5y^2-x^2/5y (putting under a common factor) remembering that 5y^2-x^2=5 in the elipse eq we can sub in for 5?, so therefore the new equation is 5/5y = 1/y?

Hint: Recall that the equation of the line can take the form y - y0 = m(x - x0) where (x0,y0) is a known point on the line.