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I tried the general method of comparing separate equations ie. $$y=mx+\frac{1}{4m}$$ and $$y-3=mx\pm \sqrt 5( \sqrt {1+m^2})$$

Then $$\frac {1}{4m} =3 \pm \sqrt {5} (\sqrt {1+m^2})$$

I got stuck in the infinite operation of squaring the both sides.

However, even if I do eventually find the value of m after an eternity, it’s still an incredibly ineffecient way to solve. Is there a way to solve it more effectively, or is there an easier if squaring operation?

Aditya
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  • The dual curves $-\frac14 Y^2+X=0$ and $-5X^2+4Y^2+6Y+1=0$ intersect in three points $(1:-2:1),(-(2\sqrt{30}-11)/5:-(2\sqrt{30}-10)/5:1),((2\sqrt{30}+11)/5:(2\sqrt{30}+10)/5:1)$ corresponding to the three common tangents: $x-2y+1=0,-\frac{2\sqrt{30}-11}{5}x-\frac{2\sqrt{30}-10}{5}y+1=0, \frac{2\sqrt{30}+11}{5}x+\frac{2\sqrt{30}+10}{5}y+1=0$ – Jan-Magnus Økland Mar 02 '20 at 14:17
  • Since you’re effectively trying to compute the intersection of two conics (the dual conics to the circle and parabola), having to solve a quartic or cubic equation seems unavoidable. – amd Mar 03 '20 at 18:58

1 Answers1

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Hint:

The equation of any tangent at $(t^2,t)$

$$\dfrac{y-t}{x-t^2}=\dfrac1{2t}\iff x-2ty+t^2=0$$

Now if this has to be a tangent of the circle as well

the distance from the center has to be $=$radius

$$\implies\sqrt5=\dfrac{|0-2t\cdot3+t^2|}{\sqrt{1+(-2t)^2}}$$

lab bhattacharjee
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  • Solving this gives me a biquadratic $t^4-12t^3+16t^2-5=0$, which is another major obstacle – Aditya Mar 02 '20 at 14:11
  • @Aditya Solving that quadric is a bit tedious, but not terribly difficult. After a couple of guesses, you can find that $1$ is a double root, leaving only a quadratic equation to solve. – amd Mar 03 '20 at 18:57