# 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 : Option 1) $(\frac{5}{2},1)$ Option 2) $(-\frac{5}{2},-1)$ Option 3) $(\frac{5}{2},-1)$ Option 4) $(-\frac{5}{2},1)$

$(x-2)^{2}-1=x-3$

$=>x=2,3$

Clearly, $A(2,-1)$  and  $B(3,0)$

Tangent to Parabola at B :

To find slope, differentiate the given curve

$\frac{dy}{dx}=2(x-2)$

$(\frac{dy}{dx})_{(3,0)}=2$

Equation of tangent at B : $y-0=2(x-3)$

=> $y=2x-6$................(1)

Equation of tangent at A : $y=-1$.........................(2)

Clearly,(1) and (2) intersect at  $(\frac{5}{2},-1)$.

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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