If 4x^2 + py^2 = 45 and x^2 - 4y^2 = 5 cut or | vedclass

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(D) For the first curve $4x^2 + py^2 = 45$,differentiating with respect to $x$ gives $8x + 2py(dy/dx) = 0$,so $(dy/dx)_I = -4x/(py)$.For the second cu

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$9$

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(D) For the first curve $4x^2 + py^2 = 45$,differentiating with respect to $x$ gives $8x + 2py(dy/dx) = 0$,so $(dy/dx)_I = -4x/(py)$.For the second curve $x^2 - 4y^2 = 5$,differentiating with respect to $x$ gives $2x - 8y(dy/dx) = 0$,so $(dy/dx)_{II} = x/(4y)$.Since the curves cut orthogonally,the product of their slopes is $-1$: $(-4x/py) 	imes (x/4y) = -1$.This simplifies to $x^2/(py^2) = 1$,or $x^2 = py^2$.Substitute $x^2 = py^2$ into the second equation: $py^2 - 4y^2 = 5$,which gives $y^2(p - 4) = 5$,so $y^2 = 5/(p - 4)$.Substitute $x^2 = py^2$ into the first equation: $4(py^2) + py^2 = 45$,which gives $5py^2 = 45$,so $py^2 = 9$.Since $y^2 = 9/p$,substitute this into $y^2 = 5/(p - 4)$ to get $9/p = 5/(p - 4)$.$9(p - 4) = 5p \implies 9p - 36 = 5p \implies 4p = 36 \implies p = 9$.

If $4x^2 + py^2 = 45$ and $x^2 - 4y^2 = 5$ cut orthogonally,then the value of $p$ is

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