Question

Q25. If 24(3x^2 - y^2) = 37xy, then x:y is
(a) 8/9 and -3/9
(b) 3/5 and 3/7
(c) 3/7 and -2/5
(d) 2/5 and -3/5​



Please give step by step answer..

Answer 1

$=>24(3x^2-y^2)=37xy$

Theory :-

In such questions, in which each term is homogeneous in $x, y$ in degree 2, then better divide the whole equation by either $x^2$ or $y^2$.

Basic :-

If you don't know the meaning of homogeneous equation then,

"Equations in which sum of powers of variable term for each term is same, is known as homogeneous equation."

Like here, Power of x in $24(3x^2)$ = 2

                Power of y in $24(-y^2)$ = 2

                Power of x, y in $37xy$ = 1 + 1 = 2

Solution :-

$=> 72x^2-24y^2=37xy$

$=>\frac{72x^2}{x^2} -\frac{24y^2}{x^2}=\frac{37xy}{x^2}$

$=>72-24(\frac{y^2}{x^2})=37(\frac{y}{x})$

Now, let $\frac{y}{x}=t$

$=>72-24t^2=37t$

Now, homogeneous equation has been reduced to a simple quadratic equation.

$=>24t^2+37t-72=0$

$=> (24t^2 +64t)+(-27t-72)=0$

$=>8t(3t+8)-9(3t+8)=0$

$=>(3t+8)(8t-9)=0$

$=> Either\ t=(\frac{-8}{3})\ or,\ t=(\frac{9}{8})$

So, $Either\ (\frac{y}{x})=(\frac{-8}{3})\ or, \ (\frac{y}{x})=(\frac{9}{8})$

So, $Either\ (\frac{x}{y})=(\frac{-3}{8})\ or, \ (\frac{x}{y})=(\frac{8}{9})$

Remarks :-

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