# Distance of point from a particular line

The number of points on the line $$3x + 4y = 5$$, which are at a distance of $$sec^2\theta+2cosec^2\theta$$ ,$$\theta \in \mathbb{R}$$ from the point $$(1, 3)$$, is
(1) 1

(2) 2

(3) 3

(4) infinite

My approach is as follow the least distance of $$3x+4y=5$$ from the point $$(1,3)$$ is $$2$$ which is perpendicular distance. The point from $$3x+4y=5$$ from the point $$(1,3)$$ may not be perpendicular. The distance $$sec^2\theta+2cosec^2\theta$$ is always greater than $$2$$ so we need to find the number of points valid for $$sec^2\theta+2cosec^2\theta$$ which I am not able to find.

• Is the answer supposed to depend on $\theta$? – Parcly Taxel Apr 14 at 7:16

The function $$\sec^2\theta+2\csc^2\theta$$ is unbounded in the positives, so there are infinitely many points that can satisfy the condition.
By C-S $$\frac{1}{\cos^2\theta}+\frac{2}{\sin^2\theta}=(\cos^2\theta+\sin^2\theta)\left(\frac{1}{\cos^2\theta}+\frac{2}{\sin^2\theta}\right)\geq(1+\sqrt2)^2,$$ which gives a range of $$\frac{1}{\cos^2\theta}+\frac{2}{\sin^2\theta}$$: $$\left[(1+\sqrt2)^2,+\infty\right)$$ and there are infinitely many such points.
• if the minimum is $(1+\sqrt{2})^2$, the range starts with it, does not it? – farruhota Apr 14 at 9:49
• should not the range be: $[(1+\sqrt{2})^2,+\infty)$? – farruhota Apr 14 at 10:26