Find the equation of the line?

mj.alawami · #1 $Old$ November 23rd, 2009, 01:06 AM

Find the equation of the line tangent to the curve $y=x^2$ and parallel to the line $x+y-2$ ?

Please answer in steps

Thank you

simplependulum · #2 $Old$ November 23rd, 2009, 01:15 AM

Quote:

Originally Posted by mj.alawami $View Post$

Find the equation of the line tangent to the curve $y=x^2$ and parallel to the line $x+y-2$ ?

Please answer in steps

Thank you

I observe that if the line is parallel to another line $x + y -2 = 0$ , which has the slope $-1$

the slope of the line is also $-1$

For this reason , there should be a point on both the line and the curve $y = x^2$

we just equate the slope $\frac{dy}{dx} = \frac{d(x^2)}{dx} = 2x = - 1 , \implies x= - \frac{1}{2}$

At $x = -\frac{1}{2}$ , $y = ( - \frac{1}{2} )^2 = \frac{1}{4}$

therefore , the point is $(-\frac{1}{2} , \frac{1}{4} )$

Now , we have a point on the line and its slope , the equation is comfirmed , it is $4x + 4y + 1 = 0$

mj.alawami · #3 $Old$ November 23rd, 2009, 01:58 AM

Quote:

Originally Posted by simplependulum $View Post$

I observe that if the line is parallel to another line $x + y -2 = 0$ , which has the slope $-1$

the slope of the line is also $-1$

For this reason , there should be a point on both the line and the curve $y = x^2$

we just equate the slope $\frac{dy}{dx} = \frac{d(x^2)}{dx} = 2x = - 1 , \implies x= - \frac{1}{2}$

At $x = -\frac{1}{2}$ , $y = ( - \frac{1}{2} )^2 = \frac{1}{4}$

therefore , the point is $(-\frac{1}{2} , \frac{1}{4} )$

Now , we have a point on the line and its slope , the equation is comfirmed , it is $4x + 4y + 1 = 0$

Can you please show me how did you convert the points into the equation please?