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Hyperbola

Section: Mathematics

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Question : 10 of 34

Marks: +1, -0

By rotating the axes about the origin in anti-clockwise direction with certain angle, if the equation

x^2+4xy+y^2=1

is transformed to

\;\frac{x^2}{a^2} - \;\frac{y^2}{b^2}=1

\sqrt{\;\frac{a^2+b^2}{a^2}} =

[AP EAPCET 22 May 2025 S1]

$2$
$\; \frac{\sqrt{13}}{3}$
$\; \frac{3}{2}$
$\sqrt{10}$

Solution:

Given equation

\;x^2+4xy+y^2=1

\;a x^2 + 2 h x y + b y^2 = 1

\;x = X \cos \theta - Y \sin \; \theta

\;y = X \sin \; \theta + Y \cos \theta

\;a=1, h=2, b=1

\;h^2 - a b = 3 > 0 \Rightarrow \; \text{ Given } \;

curve and hyperbola,

\tan 2\theta \; = \; \frac{2h}{a-b} = \; \frac{4}{0} = \; \frac{\pi}{2}

\theta \; = \; \frac{\pi}{4}

x \; = \; \frac{X-Y}{\sqrt{2}}, \; y = \; \frac{X+Y}{\sqrt{2}}

From (i), we get

\;\; \frac{X^2+Y^2-2XY}{2} + \; \frac{4(X^2-Y^2)}{2} + \; \frac{X^2+Y^2+2XY}{2} = 1

\;\; \frac{1}{2} [6 X^2 - 2 Y^2] = 1 \Rightarrow 3 X^2 - Y^2 = 1

\;\; \frac{X^2}{\frac{1}{3}} - \; \frac{Y^2}{1} = 1 \Rightarrow a^2 = \; \frac{1}{3}, b^2 = 1

\; \sqrt{\;\frac{a^2+b^2}{a^2}} = \sqrt{\;\frac{\frac{1}{3}+1}{\frac{1}{3}}} = 2

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