For the curve x^{2}+4xy+8y^{2}=64, the tangen | vedclass

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(B) Given curve is $x^{2}+4xy+8y^{2}=64$ ... $(i)$ Differentiating with respect to $x$, we get: $2x + 4(y + x\frac{dy}{dx}) + 16y\frac{dy}{dx} = 0$ $2

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$(8, -4)$ and $(-8, 4)$

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(B) Given curve is $x^{2}+4xy+8y^{2}=64$ ... $(i)$ Differentiating with respect to $x$, we get: $2x + 4(y + x\frac{dy}{dx}) + 16y\frac{dy}{dx} = 0$ $2x + 4y + (4x + 16y)\frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{x+2y}{2(x+4y)}$ Since the tangents are parallel to the $x$-axis, the slope $\frac{dy}{dx} = 0$. This implies $x + 2y = 0$, so $x = -2y$ ... (ii) Substituting $x = -2y$ into the original equation $(i)$: $(-2y)^{2} + 4(-2y)y + 8y^{2} = 64$ $4y^{2} - 8y^{2} + 8y^{2} = 64$ $4y^{2} = 64$ $y^{2} = 16 \Rightarrow y = \pm 4$ If $y = 4$, then $x = -2(4) = -8$. If $y = -4$, then $x = -2(-4) = 8$. Thus, the required points are $(-8, 4)$ and $(8, -4)$.

For the curve $x^{2}+4xy+8y^{2}=64$, the tangents are parallel to the $x$-axis only at the points

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