The point (x, y) lies on the line 2x + 3y = 6. The smallest value of the quantity is m. Then the value of 26m^2 / 5 is.

JEE MainMathematicsToughStraight Lines

The point (x, y) lies on the line 2x + 3y = 6. The smallest
value of the quantity $\sqrt{x^{2} + y^{2}}$ is m. Then the value of
26m^2 / 5 is.

Show answer and solution

Let

x = r\cos\theta;y = r\ sin\theta.

Then,

2rcos\theta + 3rsin\theta = 6

r = \frac{6}{2cos\theta + 3sin\theta}

and

r = \sqrt{x^{2} + y^{2}}

For r to be minimum,

2\ cos\ \theta + 3

sin

\theta

must be maximum,
i.e.,

\sqrt{13}

. Therefore,

r_{\min} = \frac{6}{\sqrt{13}} = m

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

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