If theta is the acute angle between the pair | vedclass

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(C) Given the equation of the pair of lines is $H \equiv ax^2 - xy + by^2 = 0$. Since the point $(1, -1)$ lies on $H = 0$,we substitute $x = 1$ and $y

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$7$

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(C) Given the equation of the pair of lines is $H \equiv ax^2 - xy + by^2 = 0$. Since the point $(1, -1)$ lies on $H = 0$,we substitute $x = 1$ and $y = -1$ into the equation: $a(1)^2 - (1)(-1) + b(-1)^2 = 0$ $a + 1 + b = 0 \Rightarrow a + b = -1$ Comparing $ax^2 - xy + by^2 = 0$ with the general form $ax^2 + 2hxy + by^2 = 0$,we get $2h = -1$,so $h = -\frac{1}{2}$. The formula for the acute angle $	heta$ between the pair of lines is $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. Given $	an 	heta = 5$,we have: $5 = \left| \frac{2\sqrt{(-\frac{1}{2})^2 - ab}}{-1} ight| = \left| \frac{2\sqrt{\frac{1}{4} - ab}}{-1} ight| = 2\sqrt{\frac{1}{4} - ab}$ Squaring both sides: $25 = 4(\frac{1}{4} - ab)$ $\Rightarrow 25 = 1 - 4ab$ $\Rightarrow 4ab = -24$ $\Rightarrow ab = -6$. Now,we need to find $a^2 + ab + b^2$. $a^2 + ab + b^2 = (a + b)^2 - ab = (-1)^2 - (-6) = 1 + 6 = 7$.

If $\theta$ is the acute angle between the pair of lines $H \equiv ax^2 - xy + by^2 = 0$,$\tan \theta = 5$ and $(1, -1)$ is a point on $H = 0$,then $a^2 + ab + b^2 =$

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