Question

If ax² + 2hxy + by² = 0, then prove that `(d^2y)/(dx^2)` = 0.

Answer 1

ax² + 2hxy + by² = 0

Differentiating w.r.t. 'x',

`2ax + 2h (x (dy)/(dx) + y.1) + 2by (dy)/(dx)` = 0

∴ `(2hx + 2by) (dy)/(dx) = - (2ax + 2hy)`

∴ `(dy)/(dx) = (-2(ax + hy))/(2(hx + by))`

∴ `(dy)/(dx) = (-(ax + hy))/((hx + by))`   ......(1)

Now, ax² + 2hxy + by² = 0

⇒ ax² + hxy + hxy + by² = 0

⇒ x(ax + hy) + y(hx + by) = 0

⇒ x(ax + hy) = – y(hx + by)

⇒ `(ax ++ hy)/(hx + by) = (-y)/x`  ......(2)

Substituting (2) in (1),

`(dy)/(dx) = - ((-y)/x) = y/x`  ......(3)

Again, differentiating w.r.t., x,

`(d^2y)/(dx^2) = (x . (dy)/(dx) - y.1)/x^2`

= `(x . y/x - y)/x^2`  ......[From (3)]

⇒ `(d^2y)/(dx^2) = (y - y)/x^2` = 0.