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

Question

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

Dr. Abubacker · Accepted Answer

(b)
The perpendicular distance of (1, 3) from the line $3x + 4y = 5$ is 2 units while,
$\sec^{2}\ 	heta + 2\ {cosec}^{2}	heta \geq 3$ [as $\sec^{2}	heta,\ {cosec}^{2}	heta \geq 1brack$
So, there will be two such points on the line

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

Explanation

Reviewed by Dr. Abubacker

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