If (a + 3b)(3a + b) = $4 h^{2}$ , then the angle between the lines represented by $a x^{2} + 2 h x y + b y^{2} = 0$ is

$60^{\circ}$

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$45^{\circ}$

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$60^{\circ}$

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${tan}^{- 1} \frac{1}{2}$

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Solution

The correct option is C
$60^{\circ}$

$θ = t a n^{- 1} (\frac{2 \sqrt{h^{2} - a b}}{a + b}) = t a n^{- 1} (\frac{\sqrt{4 h^{2} - 4 a b}}{a + b})$
= $t a n^{- 1} (\frac{\sqrt{3 a^{2} + 3 b^{2} + 10 a b - 4 a b}}{a + b}) = 60^{\circ}$ .

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If (a + 3b)(3a + b) = 4h2, then the angle between the lines represented by ax2+2hxy+by2=0 is

The correct option is C 60∘ θ=tan−1(2√h2−aba+b)=tan−1(√4h2−4aba+b) = tan−1(√3a2+3b2+10ab−4aba+b)=60∘.

If (a + 3b)(3a + b) = $4 h^{2}$ , then the angle between the lines represented by $a x^{2} + 2 h x y + b y^{2} = 0$ is

The correct option is C
$60^{\circ}$

$θ = t a n^{- 1} (\frac{2 \sqrt{h^{2} - a b}}{a + b}) = t a n^{- 1} (\frac{\sqrt{4 h^{2} - 4 a b}}{a + b})$
= $t a n^{- 1} (\frac{\sqrt{3 a^{2} + 3 b^{2} + 10 a b - 4 a b}}{a + b}) = 60^{\circ}$ .