Conic sections ( coordinates of the point of intersection of tangents with the curve ) I was told by my math teacher that equation of a pair of tangents drawn from the point is $T^2=SS_1$
This gives a second degree equation in $X$ and $Y$, which represents a pair of straight lines passing through the point from which tangent is drawn to the curve.
Is there any way I can find the coordinates of the point of intersections of the two tangents with the curve? I am very much interested in understanding conic sections.
Thank you for your help

 A: Yes this is definitely possible. This isn't the problem shown in the image, but I'll show you how to find the intersection point of $2$ tangent lines from the surface of an ellipse. We have out ellipse from the image, and let's pick $2$ points on the ellipse to use: $(1, \sqrt{32/9})$ and $(-\sqrt{135/16}, 0.5)$  First, we need to find the equations of the tangent lines we already have points, so all we need is the slopes. We can find the slopes by taking the derivative of the ellipse and solving for $\frac{dy}{dx}$:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
$$\frac{2x}{9} + \frac{2y\frac{dy}{dx}}{4} = 0$$
$$\frac{8x}{18} = -y\frac{dy}{dx}$$
$$\frac{-8x}{18y} = \frac{dy}{dx}$$
Now plug in our two points and we get a slope of $-0.2357$ and $2.582$ respectively. Now we make our two lines in point slope form and solve for $y$:
$y = -0.2357(x-1) + \sqrt{32/9}$
$y = 2.582(x+\sqrt{135/16}) + 0.5$
Then we can expand and equate the two equations to find the $x$ value for where they meet:
$2.582x + 8 = -0.2357x + 2.1213$
$x=\frac{-5.8787}{2.8177} = -2.0863$
Now plug that into one of out two tangent lines and we get an $y$ value of $2.6132$. So our two tangent lines on the ellipse intersect at a point of $(-2.0863, 2.6132)$. Here's a Desmos graph where you can see the results. I sure hope you enjoy this because it took a while to write haha.
