If theta is an acute angle between the lines | vedclass

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(C) The formula for the angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is $	an 	heta = \left| \frac{2 \sqrt{h^2 - ab}}{a + b} i

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$3$

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(C) The formula for the angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is $	an 	heta = \left| \frac{2 \sqrt{h^2 - ab}}{a + b} ight|$.Comparing $kx^2 - 4xy + y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$,we get $a = k$,$2h = -4$ (so $h = -2$),and $b = 1$.Given $	an 	heta = \frac{1}{2}$,we have $\frac{1}{2} = \left| \frac{2 \sqrt{(-2)^2 - k(1)}}{k + 1} ight|$.$\frac{1}{2} = \left| \frac{2 \sqrt{4 - k}}{k + 1} ight| \Rightarrow \frac{1}{4} = \frac{4(4 - k)}{(k + 1)^2}$.$(k + 1)^2 = 16(4 - k) \Rightarrow k^2 + 2k + 1 = 64 - 16k$.$k^2 + 18k - 63 = 0$.$(k + 21)(k - 3) = 0$.Since $	heta$ is an acute angle,the denominator $a+b = k+1$ must be such that the expression is valid. For $k = 3$,$a+b = 4 
eq 0$. Thus,$k = 3$.

If $\theta$ is an acute angle between the lines $k x^2 - 4 x y + y^2 = 0$ and $\tan \theta = \frac{1}{2}$,then the value of $k$ is

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