If two of the three lines represented by the equation $a x^{3} + b x^{2} y + c x y^{2} + d y^{3} = 0$ are perpendicular, then

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

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

Open in App

Solution

The correct option is C

The given equation being homogeneous of third degree represents three straight lines through the origin.

Since two of these lines are to be at right angles.

Let pair of these lines be $(x^{2} + p x y - y^{2})$ , p is constant and other factor is (ax - dy).

Hence, $a x^{3} + b x^{2} y + c x y^{2} + d y^{3} = (x^{2} + p x y - y^{2}) (a x - d y)$

Comparing the coefficients of similar terms, we get

b = ap - d ............(i); c = -pd - a .................(iii)

Multiplying (i) by d and (iii) by a and adding, we get

bd + ac = $- d^{2} - a^{2} \Rightarrow a^{2} + a c + b d + d^{2} = 0$ .

Suggest Corrections

Similar questions

a x^{3} + b x^{2} y + c x y^{2} + d y^{3} = 0

a^{2} + d^{2} + a c + b d

If $(a - d)^{2}$ , $a^{2}$ , $(a + d)^{2}$ are in GP, then $\frac{d^{2}}{a^{2}}$ equals to [a ≠ 0,d ≠ 0]

(a^{2} + b^{2}) x^{2} + 2 (bc + ad) x + (c^{2} + d^{2}) = 0

a^{2}, b d, c^{2}

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

(a^{2} + b^{2}) x^{2} + 2 x (a c + b d) + c^{2} + d^{2} = 0

Join BYJU'S Learning Program