Angle between pair of tangents to a conic

Question

How do u find angle between tangent to any general conic?. Well if not possible from general case then please explain in a ellipse or hyperbola.

Answer 1

To simplify the algebra, I would start by moving the point from which you are taking tangents to the origin. For example:

Find the angle between the tangents from $P=(-3,0)$ to the ellipse $x^2+4y^2=4$

1. Translate by $(3,0)$ to move $P$ to the origin. The equation of the ellipse becomes

$(x-3)^2 + 4y^2 = 4\\\Rightarrow x^2 - 6x + 4y^2 +5=0$

2. Lines through $P$ now have equation $y=mx$ so intercepts with the ellipse are solutions to

$x^2(1+4m^2) -6x + 5=0$

3. Condition for tangency is that the determinant of this quadratic is $0$ so it only has one root:

$b^2=4ac \\\Rightarrow 36 = 20(1+4m^2) \\\Rightarrow 80m^2 = 16 \\\Rightarrow m=\pm\frac{1}{\sqrt{5}}$

4. In translated co-ordinates the tangents are $y=\pm\frac{1}{\sqrt{5}}x$. The angle between them is $2\tan^{-1}\frac{1}{\sqrt{5}}$.

Answer 2

Hint:

Using $$T^2=SS_1$$ get combined equation of tangents. Now you can factor it to get equation of tangents and hence angle between them at a point gives angle between conics at that point

Answer 3

Hint

Let $y=mx+c$ be a tangent of the General conic

Replace the value of $y$ in the equation of the conic to form a quadratic equation in $x$

For tangency, the discriminant must be $0$

This will give us a quadratic equation in $m$ whose roots be $m_1,m_2$

Now we know $m_1+m_2,m_1m_2$

Can you take it from here?

Answer 4

You haven’t said so, but presumably you have a known point $P_0$ through which you’re drawing these tangents. Find the points of tangency $P_1$ and $P_2$ by intersecting the polar to $P_0$ with the conic. This is a matter of solving a couple of quadratic equations. The cosine of the angle between the lines can then be extracted from the dot product $(P_1-P_0)\cdot(P_2-P_0)$.