Two of the lines represented by the equation $a y^{4} + b x y^{3} + x x^{2} y^{2} + d x^{3} y + e x^{4} = 0$ will be perpendicular then

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

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

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

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

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Solution

The correct option is A $(b + d) (a d + b e) + {(e - a)}^{2} (a + c + e) = 0$
Let ${a y}^{4} + b x y^{3} + {c x}^{2} y^{2} + {d x}^{3} y + {e x}^{4}$
$= ({a x}^{2} + p x y - {a y}^{2}) (x^{2} + q x y + y^{2})$

Comparing the coefficients of similar terms we get,

$b = a q - p, c = - p q, d = a q + p, e = - a$

$b + d = 2 a q, e - a = - 2 a$

$a d + b e = 2 a p, a + c + e = - p q$

$(b + d) (a d + b e) = - {(e - a)}^{2} (a + c + e)$

$∴ (b + d) (a d + e b) + {(e - a)}^{2} (a + c + e) = 0$

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Similar questions

Q. Prove that two of the lines represented by the equation

a y^{4} + b x y^{3} + c x^{2} y^{2} + d x^{3} y + e x^{4} = 0

will be perpendicular if

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

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a y^{4} + b x y^{3} + c x^{2} y^{2} + d x^{3} y + w x^{4} = 0

will be at right angles if

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Two of the lines represented by the equation ay4+bxy3+xx2y2+dx3y+ex4=0 will be perpendicular then

The correct option is A (b+d)(ad+be)+(e−a)2(a+c+e)=0Let ay4+bxy3+cx2y2+dx3y+ex4=(ax2+pxy−ay2)(x2+qxy+y2)Comparing the coefficients of similar terms we get,b=aq−p,c=−pq,d=aq+p,e=−ab+d=2aq,e−a=−2aad+be=2ap,a+c+e=−pq(b+d)(ad+be)=−(e−a)2(a+c+e)∴(b+d)(ad+eb)+(e−a)2(a+c+e)=0

Two of the lines represented by the equation $a y^{4} + b x y^{3} + x x^{2} y^{2} + d x^{3} y + e x^{4} = 0$ will be perpendicular then