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  • The equation <img alt="\mathrm{a\left(x^4+y^4\right)-4 b x y\left(x^2-y^2\right)+6 c x^2 y^2= 0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba%5Cleft%28x%5E4&plus;y%5E4%5C

The equation \mathrm{a\left(x^4+y^4\right)-4 b x y\left(x^2-y^2\right)+6 c x^2 y^2= 0} represents two pairs of lines at right angles and that the two pairs will coincide, if \mathrm{2 \mathrm{~b}^2=\mathrm{a}^2+\mathrm{kac}}, where \mathrm{\mathrm{k}=}

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Given equation is \mathrm{a\left(x^4+y^4\right)-4 b x y\left(x^2-y^2\right)+ 6 c x^2 y^2=0}   .......(1)

Equation (1) is a homogeneous equation of fourth degree and since it represents two pairs at right angles. i.e., sum of the coefficients of \mathrm{x^2 \& y^2} should be zero.

Let \mathrm{a\left(x^4+y^4\right)-4 b x y\left(x^2-y^2\right)+6 c x^2 y^2=\left(a x^2+p x y-\right.\left.a y^2\right)\left(x^2+q x y-y^2\right)}
where p & q are constants.
On comparing similar powers, we get \mathrm{p+a q=-4 b}   ......(2)

\mathrm{-2 a+p q=6 c}   ......(3)

Again, given two pairs coincide, then \mathrm{\mathrm{p} / \mathrm{a}=\mathrm{q}} or \mathrm{\mathrm{p}=\mathrm{aq}}   ......(4)

from (2) \&(3), q = -2b/a and p = -2b
substituting the values of p and q in (3), we get \mathrm{-2 \mathrm{a}+4 b^2 / a=6 c}

\mathrm{\Rightarrow-a^2+2 b^2=3 a c}

Posted by

HARSH KANKARIA

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