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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.

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

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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.

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  • $\begingroup$ if the minimum is $(1+\sqrt{2})^2$, the range starts with it, does not it? $\endgroup$ – farruhota Apr 14 at 9:49
  • $\begingroup$ should not the range be: $[(1+\sqrt{2})^2,+\infty)$? $\endgroup$ – farruhota Apr 14 at 10:26
  • $\begingroup$ Oh, sorry. I'll fix it. Thank you, dear @farruhota ! $\endgroup$ – Michael Rozenberg Apr 14 at 10:28
  • $\begingroup$ no problem, good method, +1 $\endgroup$ – farruhota Apr 14 at 10:30

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