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  •  A curve  passes through the point <img alt="(6,8)" src="https://entranceco

 A curve \mathrm{y=f(x)}  passes through the point (6,8) and the normal to the curve at that point happens to be a tangent to the circle \mathrm{x^{2}+y^{2}=100} . The value of \mathrm{f^{\prime}(6)} is

Option: 1

-\frac{3}{4}


Option: 2

\frac{4}{3}


Option: 3

-\frac{4}{3}


Option: 4

\frac{3}{4}


Answers (1)

best_answer

As (6,8) lies on the circle, the normal to \mathrm{y=f(x)} is the tangent to the circle at (6,8), so that circle intersect orthogonally at (6,8)

\therefore f^{\prime}(6)=- the reciprocal of the slope of the circle at (6,8)=\frac{4}{3}

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