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Q.

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

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a

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

b

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

c

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

d

$(\frac{5}{2}, 1)$

answer is A.

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Detailed Solution

$S : {(x - 2)}^{2} - y - 1 = 0 S_{1} = (x - 2) (x_{1} - 2) - \frac{1}{2} (y + y_{1}) - 1 = 0 x (x_{1} - 2) - y \frac{1}{2} + (3 - 2 x_{1} - \frac{y_{1}}{2}) = 0 \frac{x_{1} - 2}{1} = \frac{- 1 / 2}{- 1} = \frac{3 - 2 x_{1} - \frac{y_{1}}{2}}{- 3} (x_{1}, y_{1}) = (\frac{5}{2},-1)$

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