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Trace the following central conics.
\[3{(2x - 3y + 4)^2} + 2{(3x + 2y - 5)^2} = 78.\]

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Hint: Since this equation is similar to that of ellipse hence we will compare this equation with
the standard equation of ellipse.

The standard equation of ellipse is:
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
Now comparing this equation with our given equation we see that we have an ellipse
\[3{(2x - 3y + 4)^2} + 2{(3x + 2y - 5)^2} = 78.\]
Now multiplying and dividing LHS by $13$ we have
$\dfrac{{3 \times 13{{(2x - 3y + 4)}^2}}}{{{{(\sqrt {13} )}^2}}} + \dfrac{{2 \times 13{{(3x + 2y - 5)}^2}}}{{{{(\sqrt {13} )}^2}}} = 78$
Now let $X = \dfrac{{2x - 3y + 4}}{{\sqrt {13} }},Y = \dfrac{{3x + 2y - 5}}{{\sqrt {13} }}$
on substituting it in our equation we have
$39{X^2} + 26{Y^2} = 78$
Now dividing both sides by $78$ we get
$\dfrac{{{X^2}}}{2} + \dfrac{{{Y^2}}}{3} = 1$
on comparing this equation with the standard equation of ellipse we get
$a = \sqrt 2 ,b = \sqrt 3 $

Note: While attempting question on conic sections and especially locus questions we should always compare the given equation in the question with standard equations we know of various conics because the equation in the question is always the modified equation of any of the conics and hence by comparing it and knowing of which conic it is we can further continue by then converting it to the standard form and we finally arrive to solution by this method
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