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  • Stuck here, help me understand: - Limit , continuity and differentiability - JEE Main-17

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)

 

Answers (1)

best_answer

(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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Aadil

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