Suppose that the curve has equation of √x+√y=√c. Show that the sum of the x and y intercepts of any tangent line to the curve is equal to c.
Taking the derivative of the curve implicitly we have.
12√x+dydx2√y=0dydx=−√y√x
Using Point Slope Form
y−y1=m(x−x1)y−y1=−√y1√x1(x−x1)
Recall that x-intercept is the value of x where y=0 consequently, y-intercept is the value of y where x=0. So,
For x=0,y−y1=−√y1√x1(0−x1)y=y1−√y1√x1(−x1)y=y1+√y1√x1For y=0,0−y1=−√y1√x1(x−x1)x−x1=y1√x1√y1x=y1√x1√y1+x1x=x1+√x1√y1
Thus, the x and y-intercepts are x=x1+√x1√y1 and y=y1−√x1√y1
Now, solving for the condition...
(x1+√x1√y1)+(y1√x1√y1)=cx1+2√x1√y1+y1=cbut √c=√x+√yc=(√x+√y)2c=x+√x√y+√x√y+yc=x+2√x√y+y
Therefore, it shows that the sum of the x and y-intercepts of any tangent line to the curve is equal to c.
No comments:
Post a Comment