Question

A line passing through \(P(2,3)\) and making an angle of \(30^\circ\) with the positive direction of \(x\)-axis meets \[ x^2-2xy-y^2=0 \] at \(A\) and \(B\). Then the value of \(PA\cdot PB\) is

• A. \(17\sqrt{3}+1\)
• B. \(17(\sqrt{3}+1)\)
• C. \(17(\sqrt{3}-1)\)
• D. \(17\sqrt{3}-1\)

Hint:

For a line through \((x_1,y_1)\) making angle \(\theta\), use \[ x=x_1+r\cos\theta,\qquad y=y_1+r\sin\theta \] Then substitute into the curve and use the product of roots to find \(PA\cdot PB\).

Answer 1

The Correct Option is B

Step 1: Write the parametric form of the line.
The line passes through \[ P(2,3) \] and makes an angle \[ 30^\circ \] with the positive direction of the \(x\)-axis.
A point on this line at a directed distance \(r\) from \(P\) is \[ x=2+r\cos30^\circ \] and \[ y=3+r\sin30^\circ \] Since \[ \cos30^\circ=\frac{\sqrt3}{2},\qquad \sin30^\circ=\frac12, \] we get \[ x=2+\frac{\sqrt3}{2}r \] \[ y=3+\frac12 r \]

Step 2: Substitute in the given pair of straight lines.
The given equation is \[ x^2-2xy-y^2=0 \] Substitute \[ x=2+\frac{\sqrt3}{2}r,\qquad y=3+\frac12 r \] So, \[ \left(2+\frac{\sqrt3}{2}r\right)^2 - 2\left(2+\frac{\sqrt3}{2}r\right)\left(3+\frac12 r\right) - \left(3+\frac12 r\right)^2 =0 \]

Step 3: Expand and simplify.
First, \[ x^2 = \left(2+\frac{\sqrt3}{2}r\right)^2 = 4+2\sqrt3 r+\frac34 r^2 \] Next, \[ 2xy = 2\left(2+\frac{\sqrt3}{2}r\right)\left(3+\frac12 r\right) \] \[ = 2\left(6+r+\frac{3\sqrt3}{2}r+\frac{\sqrt3}{4}r^2\right) \] \[ = 12+2r+3\sqrt3 r+\frac{\sqrt3}{2}r^2 \] Also, \[ y^2 = \left(3+\frac12 r\right)^2 = 9+3r+\frac14 r^2 \] Thus, \[ x^2-2xy-y^2=0 \] becomes \[ 4+2\sqrt3 r+\frac34 r^2 - \left(12+2r+3\sqrt3 r+\frac{\sqrt3}{2}r^2\right) - \left(9+3r+\frac14 r^2\right)=0 \] Combining like terms, \[ \left(\frac34-\frac14-\frac{\sqrt3}{2}\right)r^2 + (2\sqrt3-3\sqrt3-2-3)r + (4-12-9)=0 \] \[ \frac{1-\sqrt3}{2}r^2 + (-\sqrt3-5)r - 17=0 \]

Step 4: Use product of roots.
Let the two roots be \(r_1\) and \(r_2\), corresponding to the points \(A\) and \(B\).
Then, \[ PA\cdot PB=r_1r_2 \] For the quadratic \[ \frac{1-\sqrt3}{2}r^2+(-\sqrt3-5)r-17=0, \] product of roots is \[ r_1r_2=\frac{-17}{\frac{1-\sqrt3}{2}} \] \[ =\frac{-34}{1-\sqrt3} \] Rationalizing, \[ r_1r_2= \frac{-34(1+\sqrt3)}{(1-\sqrt3)(1+\sqrt3)} \] \[ = \frac{-34(1+\sqrt3)}{1-3} \] \[ = \frac{-34(1+\sqrt3)}{-2} \] \[ =17(1+\sqrt3) \]

Step 5: Final conclusion.
Therefore, \[ PA\cdot PB=17(\sqrt3+1) \] Hence, \[ \boxed{17(\sqrt3+1)} \] which corresponds to option (2).