analytic geometry
Let $A(-1,0),B(3,0)$ and PQ be any line passing through (4,1).The range of the slope of PQ for which there are two points on PQ at which AB subtends a right angle is $(\lambda_1,\lambda_2)$,then what is $\lambda_1+\lambda_2$.
My attempt:Let equation of PQ be $y-1=m(x-4)$.Let points on PQ at which AB subtends a right angle are $C(x_1,y_1)$and$D(x_2,y_2)$.
$\Rightarrow$$y_1-1=m(x_1-4)$ and $y_2-1=m(x_2-4)$.I could not think further steps to solve the problem.Can someone help me in this question?
There's no real problems here, we take \begin{align*} \theta &= \arctan(-\sqrt{3}) \\ &= -\arctan(\sqrt{3}) & \text{since } \arctan(x)=-\arctan(-x) \text{ for all } x \in \mathbb{R} \\ &= -\tfrac{\pi}{3}. \end{align*}
And, as we can see from the picture below, the angle should be negative:
The answer of $2\pi/3=\pi-|\theta|$ is instead the angle marked $\varphi$; this is the angle of inclination.
Hint...any point on the ellipse can be written as $$(2\cos\theta,\sin\theta)$$
The distance from this point to the line is $$\left|\frac{2\cos\theta+\sin\theta-4}{\sqrt{2}}\right|$$
Best Answer
HINT
The location of the points at which $AB$ substends a right angle is a $\color{red}{\text{circle}}$ of radius $2$ centered at $\color{red}{(1,0)}$ as shown below:
In the figure the black line through $P$ has two intersection point (green and blue) with the $\color{red}{\text{circle}}$ at the $\color{blue}{\text{point}}$ and at the $\color{green}{\text{point}}$.
The range of the straight lines form $P$ which intersect the $\color{red}{\text{circle}}$ are between the two white tangent lines. The tangent lines go through $P$ and the the intersection points of the white circle centered at $(3.5,0.5)$ and going through the center of the $\color{red}{\text{circle}}$. The center of the white circle is the midpoint of the interval joining $P$ and the center of the $\color{red}{\text{circle}}$.