Question
The number of points on the line $3x+4y=5$, which are at a distance $$(sec(\theta))^2 + 2 (cosec(\theta))^2$$, where $\theta$ ∈ $R$ , from the point (1,3) is___ .
My attempt
$$d= \frac{|3(1)+4(3)-5|}{5}$$ $$(sec(\theta))^2 + 2 (cosec(\theta))^2=|2|$$ $$(sec(\theta))^2 + 2 (cosec(\theta))^2=2$$ $$ 1+ (tan(\theta))^2 + 2+ 2(cot(\theta))^2=2$$ $$(tan(\theta))^2 + 2(cot(\theta))^2=-1$$
My doubt
It comes to be the sum of squares of two terms be negative. How is this ever possible? Is the question wrong or my thought process?
If we call the point $P$, and the line $l$, you used the distance from a $P$ to $l$. This formula gives you the smallest $d$ such that $d$ is the distance from $p$ to $L_0$, where $L_0$ is a point on $l$. That is, the distance from a point to a line is defined as the distance from the point to the closest point on the line.
The question asks you to find how many points $L_n$ there are on $l$ that are the given distance from $P$. So you want the distance from $L_n$ to $P$, not $L_0$ to $P$. The latter is a start, however. You know that the distance to the closest point is $2$. And you know that $(\sec(\theta))^2 + 2 (\csc(\theta))^2$ is at least $3$. Try drawing a graph. How do you draw the set of point of distance $3$ from $p$? How many of those points are on $l$? What if you want a distance greater than $3$?