x-intercept of tangent to parabola

**frozenflames** · January 8th 2006, 08:18 AM

Show that the tangent to the parabola Y=Ax^2 (for A does not equal 0) at the point where x = c will intersect the x axis at the point (c/2 , 0 . Where does it intersect the y axis.

**ThePerfectHacker** · January 8th 2006, 09:07 AM

Originally Posted by frozenflames

Show that the tangent to the parabola Y=Ax^2 (for A does not equal 0) at the point where x = c will intersect the x axis at the point (c/2 , 0 . Where does it intersect the y axis.

To find the tangent find the derivative of $y=ax^2$ at $x=c$ which is $\lim_{h\rightarrow 0} \frac{a(x+h)^2-ax^2}{h}=2ac$
Thus, the slope of the tangent line is $m=2ac$ . Now use the formula for finding equations of line by knowing thier slopes and the point that they pass through. That formula is $y-y_0=m(x-x_0)$ this tangent line has slope $m=2ac$ and passes through the point of tangentcy $x=c$ thus $y=ac^2$ Thus, the equation of the tangent line is $y-ac^2=2ac(x-c)$ thus, $y=2acx-2ac^2+ac^2=2acx-ac^2$ now set $y=0$ because this is the x-intercept thus $2acx-ac^2=0$ solve for $x$ thus $x=c/2$
Q.E.D.

**frozenflames** · January 8th 2006, 09:16 AM

Where does it intersect the y Axis?

**ThePerfectHacker** · January 8th 2006, 09:22 AM

Originally Posted by frozenflames

Where does it intersect the y Axis?

Easy since the equation of the tangent line is $y=2acx-ac^2$ the y-intersept is when $x=0$ thus, $y=-ac^2$
Q.E.D.

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