Question

dy
(b) Find
dx
when ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.
$​

Answer 1

Step-by-step explanation:

We have,

$a {x}^{2} + 2hxy + b {y}^{2} + 2gx + 2fy + c = 0$

Differentiating both sides, we get,

$2ax + 2hy + 2hx \frac{dy}{dx} + 2by \frac{dy}{dx} + 2g + 2f \frac{dy}{dx} + 0 = 0$

$2(ax + hy + g) +2( hx + by + f) \frac{dy}{dx} = 0$

$\frac{dy}{dx} = - 2 \frac{(ax + hy + g)}{(hx + by + f)}$