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I need help to solve for the equation of two tangent lines to the ellipse $x^2 -12x+y^2+7=0$, which pass through the origin. I've tried a variety of methods from searching similar problems on the internet, however, I can't seem to arrive at an answer.

I know from implicit differentiation that the derivative of $x^2 -12x+y^2+7=0$ with respect to $y$, is defined as:

$y' = \frac{(6-x)}{y}$

And, after that, I'm stumped. How do I go from there? Any help will be much appreciated.

Thank you in advance!

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  • $\begingroup$ If the lines pass through the origin, their equation has a very special form: $y=mx$. But you already know the slope of this tangent line at a point $(a,b)$ on the ellipse. $\endgroup$
    – MasB
    Oct 2, 2018 at 15:40

3 Answers 3

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Given the ellipse $E$ and a line passing by the origin $L$

$$ E\to b^2(x-x_0)^2+a^2(y-y_0)^2-a^2b^2 = 0\\ L\to p = (x,y) = \lambda \vec v = \lambda(1,m)\\ $$

we have that $E\cap L$ can be solved as follows

$$ b^2(\lambda-x_0)^2+a^2(\lambda m-y_0)^2-a^2b^2 = 0\ $$

for

$$ \lambda = \frac{a^2 m y_0+b^2 x_0\pm a b \sqrt{a^2 m^2+b^2-(y_0-m x_0)^2}}{a^2 m^2+b^2} $$

but at tangency

$$ a^2 m^2+b^2-(y_0-m x_0)^2 = 0 $$

and solving for $m$ we get

$$ m = \frac{\pm\sqrt{a^2 \left(y_0^2-b^2\right)+b^2 x_0^2}-x_0 y_0}{a^2-x_0^2} $$

so the tangency points are at

$$ p = \lambda_i\vec v_i\ \ \ i \in \{1,2\} $$

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  • $\begingroup$ I noticed that you used the set intersection notation. Is there a sort of application of set theory to analytic geometry and calculus? $\endgroup$
    – clathratus
    Oct 2, 2018 at 18:06
  • $\begingroup$ @clathratus I used it as a short notation for intersection. $\endgroup$
    – Cesareo
    Oct 2, 2018 at 18:14
  • $\begingroup$ Yeah I got that, I was just wondering. $\endgroup$
    – clathratus
    Oct 2, 2018 at 18:19
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First, since the coefficients of the $x^2$ and $y^2$ terms agree, if the shape the equation defines is an ellipse, it's actually a circle. In any case, you're already close to an answer.

Hint From your equation for the derivative, the tangent line to the circle at $(x_0, y_0)$ is $$y_0 (y - y_0) = (6 - x_0) (x - x_0) .$$ If this line passes through the origin, it is satisfied by $(x, y) = (0, 0)$, so $$-y_0^2 = x_0^2 - 6 x_0 .$$

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  • $\begingroup$ Thank you for the helpful hint! After getting that equation, I think I have to equate it to the original equation? $x^2 - 12x + y^2 + 7 = 0$? Then solve for the values of x? $\endgroup$
    – Ryan
    Oct 5, 2018 at 12:17
  • $\begingroup$ It's even a little easier than that: Since $(x_0, y_0)$ is on the circle, that pair satisfies the circle equation: $x_0^2 - 12 x_0 + y_0^2 + 7 = 0$. By rearranging and combining the equations, we get an affine equation in $x_0$, that is, one without quadratic terms. $\endgroup$ Oct 5, 2018 at 15:19
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This ellipse is actually a circle $$(x-6)^2+y^2=29$$ and you can find a tangent without derivative. Just write a equation of circle with diameter $A(6,0)$ and $O(0,0)$ and calculate where it cuts a given circle. Suppose you get points $B$ and $C$. Then the lines $OB$ and $OC$ are the tangnts you seek for.

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  • $\begingroup$ I assume you meant center $A(6, 0)$. $\endgroup$ Oct 3, 2018 at 12:23
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    $\begingroup$ No. I meant circle wit diameter $AO$ $\endgroup$
    – nonuser
    Oct 3, 2018 at 12:36

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