First slide

Pair of straight lines

Question

If the pair of straight lines $A x^{2} + 2 H x y + B y^{2} = 0 (H^{2} > A B)$ forms an equilateral triangle with the line $a x + b y + c = 0$ then $(A + 3 B) (3 A + B) =$

Moderate

A

$H$
B

$- H^{2}$
C

$2 H^{2}$
D

$4 H^{2}$

Solution

$\cos 60^{0} = \frac{| A + B |}{\sqrt{{(A - B)}^{2} + 4 H^{2}}} \Rightarrow {(A - B)}^{2} + 4 H^{2} = 4 {(A + B)}^{2} \Rightarrow 4 H^{2} = 3 A^{2} + 10 A B + 3 B^{2}$

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