If a x^2+2 h x y+b y^2=0, then frac{d^2 y}{d | vedclass

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(D) Given equation: $a x^2+2 h x y+b y^2=0$. Differentiating with respect to $x$: $2 a x+2 h(y+x \frac{d y}{d x})+2 b y \frac{d y}{d x}=0$. Dividing b

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$0$

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(D) Given equation: $a x^2+2 h x y+b y^2=0$. Differentiating with respect to $x$: $2 a x+2 h(y+x \frac{d y}{d x})+2 b y \frac{d y}{d x}=0$. Dividing by $2$: $a x+h y+h x \frac{d y}{d x}+b y \frac{d y}{d x}=0$. $\frac{d y}{d x}(h x+b y)=-(a x+h y) \implies \frac{d y}{d x}=-\frac{a x+h y}{h x+b y}$. Now, differentiate again with respect to $x$ using the quotient rule: $\frac{d^2 y}{d x^2} = -\frac{(h x+b y)(a+h \frac{d y}{d x})-(a x+h y)(h+b \frac{d y}{d x})}{(h x+b y)^2}$. Substituting $\frac{d y}{d x} = -\frac{a x+h y}{h x+b y}$: $\frac{d^2 y}{d x^2} = -\frac{(h x+b y)(a-h \frac{a x+h y}{h x+b y})-(a x+h y)(h-b \frac{a x+h y}{h x+b y})}{(h x+b y)^2}$. Simplifying the numerator: $\frac{d^2 y}{d x^2} = -\frac{(a h x+a b y-a h x-h^2 y)-(a h x+b h y-a b x-b h y)}{(h x+b y)^3} = -\frac{a b y-h^2 y-a h x+a b x}{(h x+b y)^3}$. Since $a x^2+2 h x y+b y^2=0$, we have $a x+h y = -\frac{h x+b y}{x} \cdot y$ (or use the property of homogeneous equations). The final result is $\frac{h^2-a b}{(h x+b y)^3} \cdot (-1)$ is not correct, the standard result for this specific homogeneous equation is $0$ because the equation represents two lines passing through the origin, so $y=mx$, thus $\frac{d^2 y}{d x^2}=0$.

If $a x^2+2 h x y+b y^2=0$, then $\frac{d^2 y}{d x^2}=$

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