Q. The length of the axes of the conic $9x^{2} + 4y^{2} - 6x + 4y + 1 = 0$, are

A) $\frac{1}{2},9\$
B) $3,\ \frac{2}{5}$
C) $1,\ \frac{2}{3}$
D) $3,2$
Explanation

(c)
Given equation is $9x^{2} + 4y^{2} - 6x + 4y + 1 = 0$
$\Rightarrow 9\left( x^{2} - \frac{2}{3}x + \frac{1}{3^{2}} \right) + 4\left( y^{2} + y + \frac{1}{4} \right) + 1 - 1 - 1 = 0$
$\Rightarrow \frac{\left( x - \frac{1}{3} \right)^{2}}{\left( \frac{1}{3} \right)^{2}} + \frac{\left( y + \frac{1}{2} \right)^{2}}{\left( \frac{1}{2} \right)^{2}} = 1$ (here, $a < b$)
Length of major axis$\ = 2b = 2\left( \frac{1}{2} \right) = 1$
Length of minor axis$\ = 2a = 2\left( \frac{1}{3} \right) = \frac{2}{3}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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