# Thread: locus- mid-pt of the chord of a ellipse

1. ## locus- mid-pt of the chord of a ellipse

i have problem in (c)
$\displaystyle (4m^2+1)x^2 +8m (1-2m)x+4(4m^2-4m -3)=0$
then i have to solve the parametric equations:

(x-coordinate of M )$\displaystyle x= -\frac{4m(1-2m)}{4m^2+1}$
and $\displaystyle y-1 = m(x-2)$
is there any other method beside put $\displaystyle m = \frac {y-1}{x-2}$ back to $\displaystyle x= -\frac{4m(1-2m)}{4m^2+1}$ ?
here's the question:

thanks!

2. Originally Posted by afeasfaerw23231233
i have problem in (c)
$\displaystyle (4m^2+1)x^2 +8m (1-2m)x+4(4m^2-4m -3)=0$
then i have to solve the parametric equations:

(x-coordinate of M )$\displaystyle x= -\frac{4m(1-2m)}{4m^2+1}$
and $\displaystyle y-1 = m(x-2)$
is there any other method beside put $\displaystyle m = \frac {y-1}{x-2}$ back to $\displaystyle x= -\frac{4m(1-2m)}{4m^2+1}$ ?
here's the question:

thanks!
Actually since P and Q both lie on the tangent as well as the ellipse, they should satisfy both the line equation and the ellipse equation.

So substitute $\displaystyle y= m(x-2)+1$ in the ellipse equation and expand to get the quadratic $\displaystyle (4m^2+1)x^2 +8m (1-2m)x+4(4m^2-4m -3)=0$. The solutions for this must be the points P and Q and thus $\displaystyle x_1$ and $\displaystyle x_2$ are the roots