# Tangent of an ellipse to an outside point

Let $C$ be a curve that is given by the equation:

$$2x^2 + y^2 = 1$$

and let P be a point $(1,1)$, which lies outside of the curve.

We want to find all lines that are tangent to $C$ and intersect $P$, and have found $y=1$, but are not sure how to find the other line.

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Write the general equation of a line that contains $P$. It should have one parameter $m$. Substitute in the equation of $C$ to get the intersection equation. Tangency means that this intersection equation (quadratic) has a double solution. This means its discriminant is $0$. This leads to a quadratic equation in the parameter $m$. You should find two solutions which is consistent with the geometric intuition. – marwalix Aug 17 '11 at 13:43

Let the equation of the line be $y = mx+c$. Since the line passes through $(1,1)$, we have $1 = m + c$ i.e. $c = 1 - m$. Hence, the equation of the line is $y = mx + 1 - m$.

Now we want the above line to be a tangent to $2x^2 + y^2 = 1$. This means the line should intersect the ellipse at exactly one point. Plug in $y = mx + 1 - m$ and the condition that the line intersects the ellipse at only one point means that the quadratic in $x$ must have only one root which means that the discriminant of the quadratic must be zero.

The quadratic we get is $2x^2 + (mx + 1 - m)^2 = 1$. Rearranging we get $$(m^2+2)x^2 + 2m(1-m)x + m^2 - 2m = 0$$ The discriminant is $4m^2(1-m)^2 - 4(m^2-2m)(m^2+2) = -4m(m-4)$

Setting this to zero, we get the two tangent from $(1,1)$ to the ellipse $2x^2 + y^2 = 1$ are $$y= 1$$ $$y = 4x - 3$$

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Thanks, that's a great answer. Does this technique work for all curves in $R^2$? – oliland Aug 17 '11 at 15:37
This idea holds good for any conic. This need not be true for arbitrary curves since for instance, the tangent to sin(x) from the point $(1,1)$ intersects the the sine curve at infinite points. – user17762 Aug 17 '11 at 15:41
You're right even if for nice curves (say $C^1$) this is valid locally which means that in a small enough interval... for algebraic curves the contact equation is a polynomial and whenever this polynomial has a root of a multiplicity higher than 1 we have tangency, the other roots are other intersection points. – marwalix Aug 18 '11 at 6:06

General equation of a line that goes through $P$ is $y-1=m(x-1)$; substituting we get $(m^2+2)x^2-2m(m-1)x+(m-1)^2-1=0$. This quadratic equation in $x$ has a double solution if and only if its discriminant is $0$. This is equivalent to $m(m-4)=0$ (do the computation).

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This is really a question in projective geometry, and has a constructive solution involving only straightedge. To see why it works, you have to know the theory, but here’s how to do it. From your outside point, call it $O$, draw two lines $\ell_1$ and $\ell_2$ each intersecting the ellipse in two points, say $\ell_1$ intersects in $A$ and $A'$, $\ell_2$ intersecting in $B$ and $B'$. Then draw the lines $\overline{AB}$ and $\overline{A'B'}$, intersecting at the point $P_1$ and the lines $\overline{AB'}$ and $\overline{A'B}$ intersecting at the point $P_2$. Then the line $m$ from $P_1$ to $P_2$ intersects your ellipse at the two points of tangency.

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Take the derivative of the equation for the ellipse: $$4x+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0$$ to get the slope, $\frac{\mathrm{d}y}{\mathrm{d}x}$, of the ellipse at $(x,y)$. The slope of a line going through $(x,y)$ and $(1,1)$ is $\frac{y-1}{x-1}$. Therefore, the points of tangency on the ellipse must satisfy $$4x+2y\frac{y-1}{x-1}=0$$

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One of the points on this tangent is (1,1) - which does not satisfy your equation? – oliland Aug 17 '11 at 16:02
The answer says "points of tangency." The point of tangency of the line $y=1$ is $(0,1)$. – André Nicolas Aug 17 '11 at 16:21