The equation $(x^{4} + y^{4}) - 4 b x y (x^{2} - y^{2}) + 6 c x^{2} y^{2} = 0$ represents two pairs of lines at right angles. The two pairs will coincide if

b^{2} = a + 3 c

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

a^{2} = b^{2} - 3 a c

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

a^{2} + b^{2} = 3 a c

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

2 b^{2} = 1 + 3 c

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

Open in App

Solution

The correct option is B $2 b^{2} = 1 + 3 c$

Suggest Corrections

Similar questions

b^{2} x^{2} - a^{2} y^{2} = 0

θ

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

The origin and the roots of the equation $x^{2} + a x + b$ = 0 form an equilateral triangle, if :

If a+b+c=0 then prove that, [(b+c)²/3bc]+[(c+a)²/3ac]+[(a+b)²/3ac]

\frac{(b^{2} + c^{2})}{3 b c} + \frac{(c + a)^{2}}{3 a c} + \frac{(a + b)^{2}}{3 a b} = \frac{1}{3}

tan θ_{1} tan θ_{2} = - \frac{a^{2}}{b^{2}}

θ_{1}

θ_{2}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

Join BYJU'S Learning Program