If the pair of lines represented by 3 x^2 - 5 | vedclass

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(D) The second equation is $6 x^2 - x y - 5 y^2 = 0$. Factoring this,we get $(6 x + 5 y)(x - y) = 0$,which gives two lines: $y = x$ and $y = -\frac{6

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$-\frac{17}{4}$

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(D) The second equation is $6 x^2 - x y - 5 y^2 = 0$. Factoring this,we get $(6 x + 5 y)(x - y) = 0$,which gives two lines: $y = x$ and $y = -\frac{6 x}{5}$. Case $1$: If $y = x$ is a common line,substitute $y = x$ into $3 x^2 - 5 x y + P y^2 = 0$: $3 x^2 - 5 x^2 + P x^2 = 0 \Rightarrow (P - 2) x^2 = 0$. For this to hold for all $x$,$P = 2$. Case $2$: If $y = -\frac{6 x}{5}$ is a common line,substitute $y = -\frac{6 x}{5}$ into $3 x^2 - 5 x y + P y^2 = 0$: $3 x^2 - 5 x(-\frac{6 x}{5}) + P(-\frac{6 x}{5})^2 = 0$ $3 x^2 + 6 x^2 + P(\frac{36 x^2}{25}) = 0$ $9 x^2 + \frac{36 P x^2}{25} = 0 \Rightarrow 9 + \frac{36 P}{25} = 0$ $36 P = -225 \Rightarrow P = -\frac{225}{36} = -\frac{25}{4}$. The sum of all possible values of $P$ is $2 + (-\frac{25}{4}) = \frac{8 - 25}{4} = -\frac{17}{4}$.

If the pair of lines represented by $3 x^2 - 5 x y + P y^2 = 0$ and $6 x^2 - x y - 5 y^2 = 0$ have one line in common,then the sum of all possible values of $P$ is

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