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Question
If `ax^2+2hxy+by^2=0` , show that `(d^2y)/(dx^2)=0`
Solution
`ax^2+2hxy+by^2=0 .........(1)`
Differentiate w.r.t. x
`2ax+2hxdy/dx+2hy+2bydy/dx=0`
`therefore ax+hxdy/dx+bydy/dx+hy=0`
`(hx+by)by/dx=-1(ax+hy)`
`therefore dy/dx=-(ax+hy)/(hx+by) ..........................(2)`
From (1),we have
`ax^2+hxy+hxy+by^2=0`
`x(ax+hy)+y(hx+by)=0`
`x(ax+hy)=-y(hx+by)`
`-(ax+hy)/(hx+by)=y/x`
Put in (2),
`dy/dx=y/x`
Differentiate w.r.t. x
`(d^2y)/(dx^2)=(xdy/dx-y)/x^2`
`=(x(y/x)-y)/x^2`
`=0/x^2`
`=0`
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