# Coordinates of the point where the normal cuts the curve

Find the equation of the normal to the curve $y= \frac{x-2}{1+2x}$ (1) at the point where the curve cuts the $x$-axis . Find the coordinates of the point where this normal cuts the curve again .

I found The equation of the normal -

$y = -5x+10$ (2)

Now , I use simultaneous equation to find the coordinates .

I sub equation 2 to 1 .

I eventually get a quadratic equation -

$-10x^2 + 14x + 12 = 0$

$x = 2 , \frac{-3}{5}$

I'm shocked now because I'm not sure which one to reject and why ?

Or do I not reject it and sub both of this x values to find 2 values of y meaning I have 2 coordinates ? Or do I have to reject ? Thanks !

• The normal cuts the curve at $(x, x-2/1+2x)$ where $x=0, -\dfrac35$ Jun 24 '17 at 13:55
• There are two intersections and you found both correctly. But "again" implies $x=-3/5$. Jun 24 '17 at 14:02

Since the normal passes through the point $(2,0)$ this is also an intersection point between the line and the curve, corresponding to your solution $x=2$. The other value of $x$ that you have found gives the other intersection.
note that the equation $$-5x+10=\frac{x-2}{1+2x}$$ factorized to $$-2\,{\frac { \left( 5\,x+3 \right) \left( x-2 \right) }{1+2\,x}}=0$$