If two curves x^2-4y^2=2 and 8x^2=40-my^2 are | vedclass

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(B) Given curves are $C_1: x^2-4y^2=2$ and $C_2: 8x^2+my^2=40$. Differentiating $C_1$ with respect to $x$: $2x - 8y \frac{dy}{dx} = 0 \implies \frac{d

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$16$

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(B) Given curves are $C_1: x^2-4y^2=2$ and $C_2: 8x^2+my^2=40$. Differentiating $C_1$ with respect to $x$: $2x - 8y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{4y} = m_1$. Differentiating $C_2$ with respect to $x$: $16x + 2my \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{8x}{my} = m_2$. Since the curves are orthogonal,$m_1 	imes m_2 = -1$. $(\frac{x}{4y}) 	imes (-\frac{8x}{my}) = -1 \implies \frac{8x^2}{4my^2} = 1 \implies 2x^2 = my^2$. From $C_1$,$x^2 = 2 + 4y^2$. Substituting this into the condition: $2(2 + 4y^2) = my^2 \implies 4 + 8y^2 = my^2 \implies (m-8)y^2 = 4$. For the curves to intersect at a point $(x, y)$,we solve the system: $x^2 - 4y^2 = 2$ and $8x^2 + my^2 = 40$. Multiply the first by $8$: $8x^2 - 32y^2 = 16$. Subtract from the second: $(m+32)y^2 = 24 \implies y^2 = \frac{24}{m+32}$. Substitute $y^2$ into $x^2 = 2 + 4y^2$: $x^2 = 2 + \frac{96}{m+32} = \frac{2m+64+96}{m+32} = \frac{2m+160}{m+32}$. Substitute into $2x^2 = my^2$: $2(\frac{2m+160}{m+32}) = m(\frac{24}{m+32}) \implies 4m + 320 = 24m \implies 20m = 320 \implies m = 16$.

If two curves $x^2-4y^2=2$ and $8x^2=40-my^2$ are orthogonal to each other,then $m=$

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