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  • A variable straight line of slope  intersects the hyperbola <img alt="\mathrm{xy=1}" src="https://entrancecorner.onc

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

Option: 1

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


Option: 2

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


Option: 3

\mathrm{16 x^2+10 x y+y^2=4}


Option: 4

None of these


Answers (1)

best_answer

Let \mathrm{P(h,k)} be any point on the locus. Equation of the line through \mathrm{P} and having slope \mathrm{4} is \mathrm{y-k=4(x-h)\ \ \ ................(i)}
Suppose this meets \mathrm{x y=1 \quad \ldots . \text { (ii) in } A\left(x_1, y_1\right) \text { and } B\left(x_2, y_2\right)}
Eliminating \mathrm{y} between \mathrm{(i)} and \mathrm{(ii)}, we get \mathrm{\frac{1}{x}-k=4(x-h)}
\mathrm{\Rightarrow \quad 1-x k=4 x^2-4 h x \quad \Rightarrow 4 x^2-(4 h-k) x-1=0\ \ ...........(iii)}
This has two roots say
 \mathrm{x_1, x_2 ; \quad x_1+x_2=\frac{4 h-k}{4}\ \ .......(iv)}\ \ \mathrm{\ and\ x_1 x_2=-\frac{1}{4}\ \ ...........(v)}
Also, \mathrm{\frac{2 x_1+x_2}{3}=h} \mathrm{\text { [ } 8 \quad P \text { divides } A B \text { in the ratio } 1: 2 \text { ] }}
\mathrm{\text { i.e., } 2 x_1+x_2=3 h\ \ \ ...............(vi)}
\mathrm{(vi) - (iv)\ gives,\ x_1=3 h-\frac{4 h-k}{4}=\frac{8 h+k}{4} \quad \text { and } \quad x_2=3 h-2 \cdot \frac{8 h+k}{4}=-\frac{2 h+k}{2}}
Putting in \mathrm{(v)} , we get \mathrm{\frac{8 h+k}{4}\left(-\frac{2 h+k}{2}\right)=-\frac{1}{4}}
\mathrm{\begin{aligned} &\mathrm{ \Rightarrow \quad(8 h+k)(2 h+k)=2 \Rightarrow 16 h^2+10 h k+k^2=2} \\ &\mathrm{ \therefore \quad \text { Required locus of } P(h, k) \text { is } 16 x^2+10 x y+y^2=2} \end{aligned}}

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