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