The two curves $y = x^{2} - 1$ and $y = 8 x - x^{2} - 9$ at the point (2, 3)

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Have a common normal 4x + y = 11

Have a common normal x + 4y–12 = 0

Have a common tangent x + 4y–14 = 0

Have a common tangent 4x–y–5 = 0

answer is A.

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Detailed Solution

$\begin{array}{rcl} y & = & x^{2} - 1 \Rightarrow \frac{dy}{dx} = 2 x = 4 \\ y & = & 8 x - x^{2} - 9 \Rightarrow \frac{dy}{dx} = 8 - 2 x = 8 - 4 = 4 \end{array}$

$∴$ eq of common tangent is $4 x - y = 5$

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