Question #495d2

1 Answer
Sep 30, 2016

Lets say that we can draw a tangent line to parabola from point P(a,b)P(a,b). Then, there exists point Q(r,s)Q(r,s) in parabola f(x)=x^2f(x)=x2 which our tangent line goes through. Because Qinf(x)Qf(x), it follows that:

f(r) = r^2 => s=r^2 => Q(r,r^2)f(r)=r2s=r2Q(r,r2)

The equation of the tangent line is: y=kx+ny=kx+n, where kk is slope and:

k=f'(x)=2x

Because our tangent line goes through Q, it follows that: k=2r and if we insert coordinates of Q and k into the equation of the tangent line, we get:

y=kx+n
r^2=2r*r+n => n = -r^2

So, our tangent line is: y=2rx-r^2.

We started with assumptation that P in y (tangent line exists and goes through P and Q). Then (insert coordinates of P into the equation of tangent line),

b=2ra-r^2 => r^2-2ra+b=0

The last equation gives us the answer: is it possible to construct the tangent line from arbitrary point P to parabola f(x)=x^2? If so, there exists the solution of the quadratic equation r^2-2ra+b=0. Because this is a quadratic equation, we may get one solution, two solutions or no solution based on the coefficients of the equation.

So, let solve it:

r^2-2ra+b=0 <=>

<=> r^2-2ra + a^2 - a^2+b=0 <=>

<=> (r-a)^2 - (sqrt(a^2-b))^2 = 0 <=>

<=> (r-a-sqrt(a^2-b))(r-a+sqrt(a^2-b))=0 <=>

<=> r-a-sqrt(a^2-b) = 0 vv r-a+sqrt(a^2-b)=0 <=>

<=> r=a+sqrt(a^2-b) vv r=a-sqrt(a^2-b)

Real solutions exist when a^2-b>=0, or b<=a^2.
When a^2=b, we get one solution:

r=a.

When a^2>b, we get two solutions:

r=a+sqrt(a^2-b)
r=a-sqrt(a^2-b)

Finally, when b>a^2 we don't have solutions.