Angle between latus rectum of hyperbola

Question

I tried it by diagram and could do nothing . Pls tell the way to solve it

Answer 1

See information of the parabolae below:

$$y^2=4b(x-2a+b) \tag{1}$$ $$\frac{dy}{dx}=\frac{2b}{y}$$

• vertex: $(2a-b,0)$
• focal length: $b$
• focus: $(2a,0)$
• semi-latus rectum: $2b$

$$x^2+4a(y-2b-a)=0 \tag{2}$$ $$\frac{dy}{dx}=-\frac{x}{2a}$$

• vertex is at $(0,a+2b)$
• focal length: $a$
• focus is at $(0,2b)$
• semi-latus rectum: $2a$

The common end of the latus rectums is the intersection $(2a,2b)$. The slopes of the tangents are $1$ and $-1$ respectively. So the meet at $90^{\circ}$ to each other.

No options equal to $90^{\circ}$. Note that $90^{\circ}=\cot^{-1} 1+\cot^{-1} 2+\cot^{-1} 3$, there may be typos.