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Question:

Find the equations of the common tangents to the parabola $y^2=2ax$ and $x^2= 2by$.

So, as we know, the equation of the tangent of $y^2=2ax$ in slope form is $y=mx+\frac{a}{2m}$; and the equation of tangent of $x^2=2by$ in slope form is $y=mx-\frac{bm^2}{2}$. Since here we are talking about the same tangent, therefore the '$m$' in both the equations is equal. So then when i just compare the two equations, the answer i get is not correct. So, can someone please point out what i have done wrong here?

Perseus
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    It would be easier to answer this if you would tell us (1) what answer you got, (2) how you got it, and (3) how you know it's not correct. – Gerry Myerson Aug 19 '21 at 05:03
  • @GerryMyerson I just directly compared the two formulas, from that i got (a/2m)= -[b(m^2)/2]. From here i am getting m^3= -a/b. I checked the answer in the solution, and it was something different. – Perseus Aug 19 '21 at 05:08
  • I just equated the two formulas because both the formulas depict the same tangents, and the 'm' in both the equations is equal since we are talking about the same common tangent and therefore the slopes of both the equations would be same. – Perseus Aug 19 '21 at 05:09
  • OK, what's the answer in the solution? – Gerry Myerson Aug 19 '21 at 05:13
  • @GerryMyerson They got m= -4a/b and 0 , but m cannot be 0 as then in the parametric form of the parabola [at^2,2at], we will get t as infinite, therefore they only took m= -4a/b. – Perseus Aug 19 '21 at 05:17
  • they solved by substituting y=mx+a/m in the equation x^2=2by, and then solved the equation by taking equal roots, which i understand, but at the same time i cannot figure out what was wrong in my idea. – Perseus Aug 19 '21 at 05:19

1 Answers1

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$m = - \sqrt[3]{\frac{a}{b}} \ $ is correct. The official solution is not correct for the stated question.

Tangent to $y^2 = 2 a x$ can be written as $y = m x + \frac{a}{2m}$

From here, we can either proceed the way you did or we can plug this in $x^2 = 2 b y$.

$x^2 = 2b (mx + \frac{a}{2m}) \implies (x-bm)^2 - b^2m^2 - \frac{ab}{m} = 0$

As the line is tangent to the parabola, we have only one solution to the equation.

So we must have, $b^2m^2 + \frac{ab}{m} = 0 \implies m = - \sqrt[3]{\frac{a}{b}} \ , $ which is same as your answer.

Math Lover
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