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Find the common tangent to: $4(x-4)^2 +25y^2 = 100$ and $4(x+1)^2 +y^2 = 4$.

I have found the derivatives of the above two equations:

$\dfrac{dy}{dx}=\dfrac{16-4x}{25y}$ and $\dfrac{dy}{dx}=\dfrac{-(4x+4)}{y}$

What do I do next?

Archer
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  • Condition such that $y=mx+c$ is tangent to an ellipse in standard form is $y=mx\pm\sqrt{a^2m^2+b^2}$. So find tangent for one ellipse and solve the other tangent with that ellipse and equate discriminant=0. – Soham Nov 28 '17 at 09:08
  • http://www.askiitians.com/forums/Analytical-Geometry/24/5789/common-tangents.htm – lab bhattacharjee Nov 28 '17 at 11:29
  • See also : https://math.stackexchange.com/questions/1270055/common-tangent-to-a-circle-and-ellipse – lab bhattacharjee Nov 28 '17 at 11:34
  • Put the equations into standard from (showing centers and semi-axes) and sketch the graphs. The answers then are obvious. There is no need to calculate the derivatives to find the answer, but they are useful to validate the answers. – Rory Daulton Nov 28 '17 at 11:51

1 Answers1

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Attempt:

1)$\dfrac{(x-4)^2}{5^2} + \dfrac{y^2}{2^2}=1;$

2) $(x+1)^2 +\dfrac{y^2}{2^2} =1;$

Draw them.

1) Major axis $5$, minor axis $2.$ Centred at $(4,0).$

2) Major axis $2$, minor axis $1.$ Centred at $(-1,0).$

The only common tangents :

1) At $(4,2)$ for ellipse $1.$

2) At $(-1,2)$ for ellipse $2.$

$y=2$;

Check your $dy/dx$ at these points.

Can you find the other common tangent?

Peter Szilas
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