
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 x2y+1=0
(note ^ symbol: is the to the power of)
If anyone could help, I would much appreciate it. Thankyou

Re: Algebraic proof
Hey Mathshelp246.
Hint: First try finding dy/dx through implicit differentiation (this corresponds to your m value).

Re: Algebraic proof
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 (Thinking)

Re: Algebraic proof
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^2x^2/5y (putting under a common factor) remembering that 5y^2x^2=5 in the elipse eq we can sub in for 5?, so therefore the new equation is 5/5y = 1/y?

Re: Algebraic proof
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.