intersection of a line and a curve

helloying · #1 $Old$ July 4th, 2009, 09:14 PM

Find the coordinates of the points of itersection A and B of the line 2x+y+2=0 and the curve 1/x +2/y= 1/2

i reform the eqn of the line to be y=-2-2x and the curve is y= $\frac{2}{(0.5-1/x)}$

then i sub the two eqn and i 2=(0.5-1/x)(-2-2x)

and so x=-2. but i only got one ans and the qn say there are two point of intersection. How to solve this qn?

artvandalay11 · #2 $Old$ July 4th, 2009, 09:51 PM

Quote:

Originally Posted by helloying $View Post$

Find the coordinates of the points of itersection A and B of the line 2x+y+2=0 and the curve 1/x +2/y= 1/2

i reform the eqn of the line to be y=-2-2x and the curve is $\frac{2}{(0.5-1/x)}$

then i sub the two eqn and i 2=(0.5-1/x)(-2-2x)

and so x=-2. but i only got one ans and the qn say there are two point of intersection. How to solve this qn?

y=-2-2x, and substitute to get $\frac{1}{x}+\frac{2}{-2-2x}=\frac{1}{2}$
$\frac{1}{x}-\frac{1}{1+x}=\frac{1}{2}$

$\frac{1+x}{x(1+x)}-\frac{x}{x(1+x)}=\frac{1}{2}$

$\frac{1}{x+x^2}=\frac{1}{2}$

$x+x^2=2$
$x^2+x-2=(x+2)(x-1)=0$

x=1 and x=-2 and therefore y=-4 and y=2

so (1,-4) and (-2,2)

malaygoel · #3 $Old$ July 4th, 2009, 11:12 PM

Quote:

Originally Posted by helloying $View Post$

2=(0.5-1/x)(-2-2x)

opening brackets,
$2=-1-x+\frac{2}{x}+2$

$2x=-x-x^2+2+2x$

which gives
$x=-2$
and $x=1$