Question

For a fixed value of $\theta$, if $n_1$ denotes the number of points on the line $3x+4y=5$ which are at a distance of $1+{\sin }^2\theta$ units from $(2,3)$ and $n_2$ denotes the number of points on $3x+4y=5$ which are at a distance of ${\sec }^2\theta +2{\text{cosec}}^2\theta$ units from $(1,3)$, then the value of $n_1+n_2$ is

Answer 1

Perpendicular distance from $(2,3)$ to the line $3x+4y-5=0$ is
$\frac{|3\times 2+4\times 3-5|}{\sqrt{3^2+4^2}}=\frac{13}{5}$
$1+{\sin }^2\theta <$ Perpendicular distance from $(2,3)$ to the line $3x+4y-5=0$
$\therefore n_1=0$

Perpendicular distance from $(1,3)$ to the line $3x+4y-5=0$ is
$\frac{|3\times 1+4\times 3-5|}{\sqrt{3^2+4^2}}=2$
Now, ${\sec }^2\theta +2{\text{cosec}}^2\theta$
$=3+{\tan }^2\theta +2{\cot }^2\theta$
$\ge 3+2\sqrt{2}$ (Using AM > GM)
${\sec }^2\theta +2{\text{cosec}}^2\theta >$ Perpendicular distance from $(1,3)$ to the line $3x+4y-5=0$
$\therefore n_2=2$