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,…

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
  1. $\frac{33}{4}$
  2. $\frac{17}{4}$
  3. $-\frac{33}{4}$
  4. $-\frac{17}{4}$

Solution

$6 x^2-x y-5 y^2=0$ $\Rightarrow(6 x+5 y)(x-y)=0 \Rightarrow y=x$ or $y=\frac{-6 x}{5}$ When $y=x, 3 x^2-5 x^2+P x^2=0 \Rightarrow P=2$
When $y=-\frac{6 x}{5} ; 3 x^2-5 x\left(\frac{-6 x}{5}\right)+\frac{36 x^2}{25} P=0$
$\Rightarrow 9+\frac{36 P}{25}=0 \Rightarrow P=\frac{-25}{4}$;
Sum of values of $P=2-\frac{25}{4}=\frac{-17}{4}$.

Asked in: AP EAMCET 2024 (21 May Shift 2)

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