Question:

AB is a focal chord of $x^2 - 2x + y - 2 = 0$ whose focus is S. If $AS = l_1$. then BS is equal to

Updated On: Jul 10, 2024
  • $l_1$
  • $\frac{4l_{1}}{4l_{1} -1}$
  • $\frac{l_{1}}{4l_{1} -1}$
  • $\frac{2l_{1}}{4l_{1} -1}$
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The Correct Option is C

Solution and Explanation

Given curve is $x^2 - 2x + y - 2 = 0$ $\Rightarrow x^{2} - 2x + y - 2 +1 = 1$ $\Rightarrow \left(x -1\right)^{2} = - y + 3 = -1\left( y - 3\right)$ which is downward parabola with $ a = \frac{1}{4}$ We know, if $l_{1}$ and $l_{2}$ are the length of the segment of any focal chord then length of semi-latus rectum is $\frac{2l_{1}l_{2}}{l_{1}+l_{2}}$ Here $AS =l_{1}$ and $BS = l_{2}$ (say) are the segments. $\therefore\quad$ we have $\frac{2l_{1}\left(BS\right)}{l_{1}+BS} = 2a \Rightarrow BS = \frac{l_{1}}{4l_{1} - 1}$
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