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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D

We know that the pair of lines

$(a^{2} - 3 b^{2}) x^{2} + 8 a b x y + (b^{2} - 3 a^{2}) y^{2} = 0$ with the line ax + by + c = 0 from an equilateral triangle.Hence Comparing with $A x^{2} + 2 H x y + B y^{2} = 0$ , then

$A = a^{2} - 3 b^{2}, B = b^{2} - 3 a^{2}, 2 H = 8 a b$ .

Now, (A + 3B)(3A + B) = $(- 8 a^{2}) (- 8 b^{2})$

= $(8 a b)^{2} = (2 H)^{2} = 4 H^{2}$

Suggest Corrections

Similar questions

A x^{2} + 2 H x y + B y^{2} = 0 (H^{2} > A B)

a x + b y + c = 0

(A + 3 B) (3 A + B) =

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

If the lines given by ax2 +2hxy + by2 =0from an equilateral triangle with the lines Lx+my =1then show that (3a+b)(a+3b)-4h2=0

If the equation $a x^{2} + 2 h x y + b y^{2} = 0$ has one line as the bisector of angle between the coordinate axes, then

a x^{2} + 2 h x y + b y^{2} = 0

(a - b)^{2} = 4 h^{2}

Join BYJU'S Learning Program