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  • If two of the lines represented by <img alt="\mathrm{ a x^4+b x^3 y+c x^2 y^2+d x y^3+a y^4=0 }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20a%20x%5E4&plus;b%20x%5E3%20y

If two of the lines represented by \mathrm{ a x^4+b x^3 y+c x^2 y^2+d x y^3+a y^4=0 } bisects the angle between the other two, then \mathrm{ \mathrm{c}+6 \mathrm{a}=\mathrm{k}_1 \: and\: \mathrm{b}+\mathrm{d}=\mathrm{k}_2 }, then \mathrm{ \mathrm{k}_1 \: and \: \mathrm{k}_2 } are respectively:
 

Option: 1

1,0


Option: 2

0,1


Option: 3

1,1


Option: 4

0,0


Answers (1)

best_answer

We know that bisectors are mutually perpendicular to each other then \mathrm{a x^4+b x^3 y+c x^2 y^2+d x y^3+a y^4=0} represents two pairs of mutually perpendicular lines
Let \mathrm{a x^4+b x^3 y+c x^2 y^2+d x y^3+a y^4=\left(a x^2+p x y-a y^2\right)\left(x^2+q x y-y^2\right)} where \mathrm{ p \& q} are constants.
Comparing the coefficients of similar terms, we get.
\begin{aligned} & \mathrm{b}=\mathrm{aq}+\mathrm{p} \\ &\end{aligned}                           ...........(i)

\begin{aligned} \mathrm{c}=\mathrm{pq}-2 \mathrm{a} \\ \\ &\end{aligned}                        ............(ii)

\begin{aligned} & \mathrm{d}=\mathrm{p}-\mathrm{aq} \\ &\end{aligned}                         ............(iii)
adding (i) and (iii), we have
\begin{aligned} & \mathrm{b}+\mathrm{d}=0 \\ & \mathrm{pq}=\mathrm{c}+2 \mathrm{a} \end{aligned}
From (ii)
But given that bisector of one pair are given by the other, i.e. ,

\mathrm{\frac{x^2-y^2}{a-(-a)}=\frac{x y}{(p / 2)} }
\mathrm{\Rightarrow \quad \frac{x^2-y^2}{4 x y}=\frac{a}{p}}                     ..........(v)
is the same as the other pair
\mathrm{ x^2+q x y-y^2=0 }
or
\mathrm{ \frac{x^2-y^2}{4 x y}=-\frac{q}{4} }               ..........(vi)
from (v) and (vi)
\mathrm{ \begin{aligned} \frac{\mathrm{a}}{\mathrm{p}} & =-\frac{\mathrm{q}}{4} \\ \therefore \quad \mathrm{pq} & =-4 \mathrm{a} \end{aligned} }
Again from (iv)
\mathrm{ \begin{aligned} & -4 \mathrm{a}=\mathrm{c}+2 \mathrm{a} \\ & \therefore \quad c+6 a=0 \\ & \end{aligned} }

Hence option 4 is correct.

 

Posted by

avinash.dongre

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