The circle x^2 +(y-c)^2 =r^2, where {co, ro}, lies inside the parabola y=x^2. The circle touches the parabola at exactly 2 points on symmetrically opposite sides of of the y-axis.

Show that 4c=1+ 4r^2

Question

The circle x^2 +(y-c)^2 =r^2, where {c>o, r>o}, lies inside the parabola y=x^2. The circle touches the parabola at exactly 2 points on symmetrically opposite sides of of the y-axis.

Show that 4c=1+ 4r^2

anonymous · Answer

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sirm3d · Answer

is this from a calculus course?

anonymous · Answer

Nope, simply from a high school exam (no ap calc or anything either)

anonymous · Answer

Been trying to figure out the equation for the point at which the circle touches parabola, think im missing something obvious lol

sirm3d · Answer

i'll try this without calculus. minus the calculus, this may be a difficult question.

sirm3d · Answer

just solve the system $$\begin{cases} y=x^2\ x^2+(y-c)^2=r^2 \end{cases}$$ for the variable $y$, then argue that the two values of $y$ must be the same, concluding $$4c=1+4r^2$$

sirm3d · Answer

the system $$\begin{cases}y=x^2\x^2+(y-c)^2=r^2\end{cases}$$ yields the equation $$y+(y-c)^2=r^2$$Expanding ang completing squares, $$\left( y+\frac{ 1-2c }{ 2 } \right)^2=r^2+\frac{ 1 }{ 4 }-c$$since the curves intersect at exactly one $y-$value, we conclude that $$r^2+\frac{1}{4}-c=0$$or $$4r^2+1=4c$$

anonymous · Answer

Nice work, hadn't considered the effect of completing the square.
Why does the LHS become 0 when there is only one y-value?

sirm3d · Answer

as for the second part, if $$4r^2+1=4c$$then $$\left(y+\frac{1-2c}{2}\right)^2=0$$or $$y=c-\frac{1}{2}$$ which is the $y$-coordinate of the points of intersection. because the points of intersection lie above the horizontal axis, we have $y>0$ or $$c-\frac{1}{2}>0$$or $$c>\frac{1}{2}$$ which is what is to be deduced.

sirm3d · Answer

|dw:1355549284791:dw|this is the case when there are two different solutions. the quadratic equation in $y$ is $$y=-\frac{1}{2}+c \pm \sqrt{r^2+\frac{1}{4}-c}$$

sirm3d · Answer

the quadratic equation $$\left(y+\frac{1-2c}{2}\right)^2=r^2+\frac{1}{4}-c$$ has $$\begin{cases}2 \text{ answers} & r^2+\frac{1}{4}-c>0\1 \text{ answer} & r^2+\frac{1}{4}-c=0\ 0 \text{ answer} & r^2+\frac{1}{4}-c<0\end{cases}$$

anonymous · Answer

Ooh I see, makes sense. Thanks for the answer, question been bugging me for the last few days lol