"A parabola with vertex (0,1) is touching the x-axis , what is the minimum integral distance of the point of contact of parabola with the x-axis from the origin"
My attempt :
let the point of contact of parabola with x axis be (h,0) , and the slope of axis of the parabola be m So Equation of Axis is : y-1=mx
Now using reflection property of a parabola , a line with slope "-m" and passing through the point (h,0) would meet the axis at focus
So I found the co-ordinates of focus as ($h\over 2$-$1\over 2m$ , $mh\over 2$+$1\over 2$)
But I am stuck on how to go about finding a constraint in order to find the minimum positive value of h
Could someone please help me to proceed further ?
Thanks in advance
I’m not sure that knowing where the focus is really helps here. The key, I think, is the tangency condition.
If the equation of the axis is $mx-y+1=0$, then an equation of the tangent at the vertex $(0,1)$ is $x+my-m=0$, therefore an equation of a parabola with vertex at $(0,1)$ has the form $$(mx-y+1)^2=p(x+my-m).\tag1$$ We can get away with using a slope for the axis because a parabola with a vertical axis obviously isn’t possible here.
If $(h,0)$ lies on this parabola, we have $$(mh+1)^2=p(h-m).\tag2$$ The tangent to the parabola at this point has an equation of the form $$(mh+1)(mx-y+1)=\frac p2(x+my-2m+h)$$ which can be rearranged into $$\left(m+m^2h-\frac p2\right)x-\left(hm+\frac12mp+1\right)y+\left(hm+mp-\frac12hp+1\right)=0.\tag3$$ From equation (2) we have $p={(mh+1)^2\over h-m}$. Both the coefficient of $x$ and the constant term of equation (3) must vanish. Substituting into the former and factoring yields the condition $${(mh+1)(2m^2-mh+1)\over2(m-h)}=0.\tag4$$ Solving for $m$ in terms of $h$ and discarding the degenerate solution will give you a constraint on $h$. If you start instead with the constant term of equation (3) you will get one more factor than (4) in the resulting condition, but that additional factor has no real roots.