If the pair of lines 3x^2 - 5xy + py^2 = 0 an | vedclass

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(C) The second equation is $6x^2 - xy - 5y^2 = 0$. Factoring this,we get $(6x + 5y)(x - y) = 0$. So the lines are $y = -\frac{6}{5}x$ and $y = x$. If

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$2, -\frac{25}{4}$

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(C) The second equation is $6x^2 - xy - 5y^2 = 0$. Factoring this,we get $(6x + 5y)(x - y) = 0$. So the lines are $y = -\frac{6}{5}x$ and $y = x$. If $y = x$ is a common line,it must satisfy $3x^2 - 5x(x) + p(x)^2 = 0$,which gives $3x^2 - 5x^2 + px^2 = 0$,so $px^2 = 2x^2$,implying $p = 2$. If $y = -\frac{6}{5}x$ is a common line,it must satisfy $3x^2 - 5x(-\frac{6}{5}x) + p(-\frac{6}{5}x)^2 = 0$. This simplifies to $3x^2 + 6x^2 + p(\frac{36}{25})x^2 = 0$,which gives $9x^2 + \frac{36p}{25}x^2 = 0$. Dividing by $x^2$,we get $9 + \frac{36p}{25} = 0$,so $\frac{36p}{25} = -9$,which means $p = -9 	imes \frac{25}{36} = -\frac{25}{4}$. Thus,$p = 2$ or $p = -\frac{25}{4}$.

If the pair of lines $3x^2 - 5xy + py^2 = 0$ and $6x^2 - xy - 5y^2 = 0$ have one line in common,then $p =$

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