Line passing through the point P (2, 3) meets the lines represented by x2−2xy−y2=0 at the points A and B such that PA.PB=17, the equation of the line is

# Line passing through the point  meets the lines represented by${x}^{2}-2xy-{y}^{2}=0$ at the points A and B such that $PA.PB=17$, the equation of the line is

1. A

$x=2$

2. B

$y=3$

3. C

$3x-2y=0$

4. D

none of these

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### Solution:

Let the equation of the line through making an angle $\theta$ with the positive direction of x-axis be

$\frac{x-2}{\mathrm{cos}\theta }=\frac{y-3}{\mathrm{sin}\theta }$.

Then the coordinates of any point on this line at a distance

r from p are $\left(2+r\mathrm{cos}\theta ,3+r\mathrm{sin}\theta \right)$. If $PA={r}_{1}$ and $PB={r}_{2}$, then ${r}_{1},{r}_{2}$ are the roots of the equation.

$\begin{array}{l}\left(2+r\mathrm{cos}\theta {\right)}^{2}-2\left(2+r\mathrm{cos}\theta \right)\left(3+r\mathrm{sin}\theta \right)-\left(3+r\mathrm{sin}\theta {\right)}^{2}=0\\ ⇒{r}^{2}\left(\mathrm{cos}2\theta -\mathrm{sin}2\theta \right)-2r\left(\mathrm{cos}\theta +5\mathrm{sin}\theta \right)-17=0\end{array}$

$⇒17=\mathbf{PA}\cdot \mathbf{PB}={r}_{1}{r}_{2}=\frac{17}{\mathrm{cos}2\theta -\mathrm{sin}2\theta }$

$⇒\mathrm{cos}2\theta -\mathrm{sin}2\theta =1$ which is satisfied by and thus
the equation of the line is.  +91

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