Question

how to find the transformation equation of x²+2√3xy-y²=2a² when the axes are rotated through an angle 30°

Answer 1

SOLUTION :

TO DETERMINE

The transformed equation of

$\sf{ {x}^{2} + 2 \sqrt{3} \: xy - {y}^{2} \: = 2 {a}^{2} }$

when the axes are rotated through an angle 30°

EVALUATION

Let P(x, y) be the coordinate of a point in the old coordinate system xy.

Now the axes are rotated through an angle 30°

So that the system transformed into coordinate system

Let P(x', y') be the coordinates of the point P in the new system x'y'

Then the transformation formula is

$\displaystyle \sf{ x = x' \cos {30}^{ \circ} - y' \sin {30}^{ \circ} \: }$

$\therefore \displaystyle \sf{ \: \: x = \frac{1}{2} ( \: x' \sqrt{3} - y' \: ) \: }$

And

$\displaystyle \sf{ y = x' \sin {30}^{ \circ} + y' \cos {30}^{ \circ} \: }$

$\therefore \displaystyle \sf{ \: \: y = \frac{1}{2} ( \: x' + y' \sqrt{3} \: ) \: }$

Now the given equation is

$\sf{ {x}^{2} + 2 \sqrt{3} \: xy - {y}^{2} \: = 2 {a}^{2} }$

Which transformed into

$\displaystyle\sf{ \frac{1}{4} {( \: x' \sqrt{3} - y')}^{2} + 2 \sqrt{3} \times \frac{1}{2} ( x' \sqrt{3} - y' ) \times \frac{1}{2} ( x' + y' \sqrt{3} ) - \frac{1}{4} {( \: x' + y' \sqrt{3} \: )}^{2} \: = 2 {a}^{2} }$

On simplification

$\implies \sf{2( \: {x' \: }^{2} - {y' \: }^{2} ) = 2{a}^{2} }$

$\implies \sf{ {x' \: }^{2} - {y' \: }^{2} = {a}^{2} }$

RESULT

Hence the transformed equation is

$\sf{{ {x' \: }^{2} - {y' \: }^{2} = {a}^{2} }}$

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LEARN MORE FROM BRAINLY

The coordinates (x,y) of a particle moving along a curve at any time t are given by

y'+ 2x = sin 2t , x' -2 y = cos 2t

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