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.
 A: 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.
A: 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).
A: 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
$$y= 1$$ $$y = 4x - 3$$
A: 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
$$
4x+2y\frac{y-1}{x-1}=0
$$
