Question

If $a{x^2} + 2hxy + b{y^2} = 0$, then $\dfrac{{dy}}{{dx}}$ is:

Answer 1

We are going to solve this problem by differentiating a given expression with respect to variable x. Using the formula of differentiation we will get the desired answer. And the formula is given below.

Formula used:
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
$d(xy) = d(y)x + d(x)y$

Complete step-by-step answer:
First we are going to differentiate the given equation in the problem with respect to x,
$a{x^2} + 2hxy + b{y^2} = 0$Where a, b and h are constants.
Now, we are going to differentiate both sides with respect to x
We are going to use the differentiation formula.
Now we are going to differentiate $a{x^2}$with respect to x. Since a is constant, we can only differentiate the term ${x^2}$and fix a.
$\Rightarrow \dfrac{d}{{dx}}\left( {a{x^2}} \right) = a\dfrac{d}{{dx}}\left( {{x^2}} \right) = 2ax$
Now we are going to differentiate $2hxy$with respect to x. Since h is constant, we can only differentiate the terms$x$, y and fix h.
$\Rightarrow \dfrac{d}{{dx}}\left( {2hxy} \right) = 2h\dfrac{d}{{dx}}\left( {xy} \right) = 2hd(xy)$
Now we are going to differentiate $b{y^2}$with respect to x. Since b is constant, we can only differentiate the term ${y^2}$and fix b.
$\Rightarrow \dfrac{d}{{dx}}(b{y^2}) = b\dfrac{d}{{dx}}({y^2}) = 2by\dfrac{{dy}}{{dx}}$
Now we are going to add the above 3 equations.
$\Rightarrow 2ax + 2hd(xy) + 2by\dfrac{{dy}}{{dx}} = 0$---- [1]
Now we use product rule as stated above to find out the differentiation of $d(xy)$ in equation [1]
$d(xy) = d(x) \times y + x \times d(y)$
We apply this rule to equation [1]
$\Rightarrow 2ax + 2h(x\dfrac{{dy}}{{dx}} + y) + 2by\dfrac{{dy}}{{dx}} = 0$
Now we are going to keep terms $\dfrac{{dy}}{{dx}}$ on one side and the remaining terms on the other side.
$\Rightarrow (2hx\dfrac{{dy}}{{dx}} + 2by\dfrac{{dy}}{{dx}}) = - 2ax - 2hy$
Now we take $\dfrac{{dy}}{{dx}}$ commonly on the left hand side term.
$\Rightarrow \dfrac{{dy}}{{dx}}(2hx + 2by) = - 2ax - 2hy$
Now, we take minus common from right hand side from above equation we get,
$\dfrac{{dy}}{{dx}}(2hx + 2by) = - \left( {2ax + 2hy} \right)$
Now we are going to divide both sides by 2 we get,
$\Rightarrow \dfrac{{dy}}{{dx}}(hx + by) = - \left( {ax + hy} \right)$
Now we are going to divide the above equation by $hx + by$ on both sides.
$\dfrac{{dy}}{{dx}} = \dfrac{y}{x}$ $\dfrac{{dy}}{{dx}} = - \dfrac{{(ax + hy)}}{{(hx + by)}}$
Now we multiply numerator and denominator by $xy$ to above expression we get
$\dfrac{{dy}}{{dx}} = - \dfrac{{(ax + hy)}}{{(hx + by)}} \times \dfrac{{xy}}{{xy}}$
Now we multiply $(ax + hy)$ by $x$in numerator and $(hx + by)$ by y in denominator we get the below equation,
$\dfrac{{dy}}{{dx}} = - \dfrac{y}{x} \times \dfrac{{(a{x^2} + hxy)}}{{(hxy + b{y^2})}}$ ---------- [2]
Now, from given question we can write,
$a{x^2} + 2hxy + b{y^2} = 0$
$\Rightarrow a{x^2} = - 2hxy - b{y^2}$
Now we put above equation in equation [2] we get,
$\dfrac{{dy}}{{dx}} = - \dfrac{y}{x} \times \dfrac{{( - b{y^2} - 2hxy + hxy)}}{{(hxy + b{y^2})}}$
Now we add -2hxy and hxy we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{y}{x} \times \dfrac{{( - b{y^2} - hxy)}}{{(hxy + b{y^2})}}$
Now we take minus commonly from right hand side
$\Rightarrow \dfrac{{dy}}{{dx}} = - ( - )\dfrac{y}{x} \times \dfrac{{(b{y^2} + hxy)}}{{(hxy + b{y^2})}}$
Now multiplying (-) by (-) we get (+) sign and canceling numerator and denominator by $b{y^2} + hxy$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x}$
Hence, $\dfrac{{dy}}{{dx}} = \dfrac{y}{x}$

Note:
We have to use the proper formula for differentiating equations and we have to be careful of the dependent and independent variables while differentiating using product rule.
By seeing the question, we must figure out what method and which formula should be used to get the solution easily. In this problem we could not separate y and x easily so we have to differentiate at the initial step.