If the curves y^2=16x and 9x^2+alpha y^2=25 i | vedclass

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(C) Let the curves be $C_1: y^2 = 16x$ and $C_2: 9x^2 + \alpha y^2 = 25$. First,find the point of intersection $(x_1, y_1)$. From $C_1$,$y^2 = 16x$. S

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$\frac{9}{2}$

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(C) Let the curves be $C_1: y^2 = 16x$ and $C_2: 9x^2 + \alpha y^2 = 25$. First,find the point of intersection $(x_1, y_1)$. From $C_1$,$y^2 = 16x$. Substituting this into $C_2$: $9x^2 + \alpha(16x) = 25 \implies 9x^2 + 16\alpha x - 25 = 0$. For the curves to intersect at right angles,the product of their slopes at the point of intersection must be $-1$. Differentiating $C_1$: $2y \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = \frac{8}{y}$. Differentiating $C_2$: $18x + 2\alpha y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{9x}{\alpha y}$. At the point of intersection,the product of slopes is $(\frac{8}{y})(-\frac{9x}{\alpha y}) = -1 \implies \frac{72x}{\alpha y^2} = 1$. Since $y^2 = 16x$,we have $\frac{72x}{\alpha(16x)} = 1 \implies \frac{72}{16\alpha} = 1 \implies 16\alpha = 72 \implies \alpha = \frac{72}{16} = \frac{9}{2}$.

If the curves $y^2=16x$ and $9x^2+\alpha y^2=25$ intersect at right angles,then $\alpha=$

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