logo

Ask Questions, Get Answers

X
 

If $ax^2+2hxy+by^2+2gx+2fy+c=0$,then show that $\large \frac{dy}{dx}.\frac{dx}{dy}\normalsize =1$

Download clay6 mobile app

1 Answer

Toolbox:
  • A function $f(x,y)$ is said to be implicit if it is jumbled in such a way,that it is not possible to write $y$ exclusively s a function of $x$.
  • $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx}$
Step 1:
$ax^2+2hxy+by^2+2gx+2fy+c=0$
Differentiating w.r.t $x$ we get,
Apply product rule to differentiate $xy$
$2ax+2h[x.\large\frac{dy}{dx}$$+y.1]+2by.\large\frac{dy}{dx}$$+2g+2f.\large\frac{dy}{dx}$
$\Rightarrow \large\frac{dy}{dx}$$[2hx+2by+2f]=-(2ax+2hy+2g)$
$\large\frac{dy}{dx}$$=-\large\frac{(2ax+2hy+2g)}{2hx+2by+2f}$
Step 2:
Differentiating w.r.t $y$ we get,
$2ax.\large\frac{dx}{dy}$$+2h[\large\frac{dx}{dy}$$.y+1.x]+2by+2g.\large\frac{dx}{dy}$$+2f=0$
Therefore $\large\frac{dx}{dy}$$[2ax+2hy+2g]=-(2hx+2by+2f]$
$\Rightarrow \large\frac{dx}{dy}=\bigg[\large\frac{-(2hx+2by+2f)}{2ax+2hy+2g}\bigg]$
Therefore $\large\frac{dy}{dx}$$\times \large\frac{dx}{dy}=\large\frac{-(2ax+2hy+2g)}{2hx+2by+2f}$$\times \large\frac{(-2hx+2by+2f)}{(2ax+2hy+2g)}$
$\qquad\qquad\qquad\;\;\;=1$
Hence proved.
answered Jul 2, 2013 by sreemathi.v
 
...
X