The tangents to the curve

Question:

The tangents to the curve $y=(x-2)^{2}-1$ at its points of intersection with the line $x-y=3$, intersect at the point :

  1. $\left(-\frac{5}{2},-1\right)$

  2. $\left(-\frac{5}{2}, 1\right)$

  3. $\left(\frac{5}{2},-1\right)$

  4. $\left(\frac{5}{2}, 1\right)$


Correct Option: , 3

Solution:

Put $\mathrm{x}-2=\mathrm{X} \& \mathrm{y}+1=\mathrm{Y}$

$\therefore$ given curve becomes $\mathrm{Y}=\mathrm{X}^{2}$ and $\mathrm{Y}=\mathrm{X}$

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