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A variable straight line of slope \mathrm{4} intersects the hyperbola \mathrm{xy=1} at two points. Find the locus of the point which divides the line segment between these two points in the ratio \mathrm{1:2.}

 

Option: 1

\mathrm{4 x^2+8 x y+y^2-1=0}


Option: 2

\mathrm{16 x^2+10 x y+y^2-2=0}


Option: 3

\mathrm{4 x^2+8 x y+y^2-2=0}


Option: 4

\mathrm{16 x^2+10 x y+y^2-1=0}


Answers (1)

best_answer

Let equation of the line be \mathrm{y=4x+c} where \mathrm{c} is parameter. It intersects the hyperbola \mathrm{xy=1} at two points, for which \mathrm{x(4 x+c)=1 \Rightarrow 4 x^2+c x-1=0}.
Let \mathrm{x_{1}} and \mathrm{x_{2}} be the roots of this equation. Then \mathrm{x_1+x_2=-c / 4 \text\ { and }\ x_1 x_2=-1 / 4}
If \mathrm{A} and \mathrm{B} are the points of intersection of the line and the hyperbola, then the co-ordinates of \mathrm{A} are \mathrm{(x_{1},1/x_{1})} and that of \mathrm{B} are \mathrm{(x_{2},1/x_{2})}
Let \mathrm{R\left ( h,k \right )} be the point which divides \mathrm{AB} in the ratio \mathrm{1:2,} then 
\mathrm{\mathrm{h}=\frac{2 \mathrm{x}_1+\mathrm{x}_2}{3} \text { and } \mathrm{k}=\frac{2 / \mathrm{x}_1+1 / \mathrm{x}_2}{3}=\frac{2 \mathrm{x}_2+\mathrm{x}_1}{3 \mathrm{x}_1 \mathrm{x}_2}}          \mathrm{\Rightarrow 2 \mathrm{x}_1+\mathrm{x}_2=3 \mathrm{~h} \ldots \ldots \text { (i) }}
and \mathrm{\mathrm{x}_1+2 \mathrm{x}_2=3(-1 / 4) \mathrm{k}=(-3 / 4) \mathrm{k} \ldots \ldots . \text { (ii) }}
adding \mathrm{\left ( i \right )} and \mathrm{\left ( ii \right )} we get  \mathrm{3\left(\mathrm{x}_1+\mathrm{x}_2\right)=3[\mathrm{~h}-\mathrm{k} / 4]}
\mathrm{3\left(-\frac{c}{4}\right)=3\left(h-\frac{k}{4}\right) \Rightarrow h-\frac{k}{4}=-\frac{c}{4}}       \mathrm{.......\left ( iii \right )}
subtracting \mathrm{\left ( ii \right )} from \mathrm{\left ( i \right )} we get. \mathrm{x_{1}-x_{2}=3\left ( h+k/4 \right )}
\mathrm{\Rightarrow\left(\mathrm{x}_1-\mathrm{x}_2\right)^2=9(\mathrm{~h}+\mathrm{k} / 4)^2}             \mathrm{\Rightarrow \frac{\mathrm{c}^2}{16}+1=9(\mathrm{~h}+\mathrm{k} / 4)_2}
\mathrm{\Rightarrow(\mathrm{h}-\mathrm{k} / 4)^2+1=9\left(\mathrm{~h}^2+\frac{1}{2} \mathrm{hk}+\frac{\mathrm{k}^2}{16}\right)}            \mathrm{\Rightarrow 16 \mathrm{~h}^2+10 \mathrm{hk}+\mathrm{k}^2-2=0}
So that the locus of \mathrm{\mathrm{R}(\mathrm{h}, \mathrm{k}) \text { is } 16 \mathrm{x}^2+10 \mathrm{xy}+\mathrm{y}^2-2=0}

 

Posted by

manish painkra

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