Tangents and normals

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$The tangent to the curve y = e^{x} drawn at the point (c, e^{c}) intersects the$ $line joining the points (c - 1, e^{c - 1}) and (c + 1, e^{c + 1})$

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Solution

$\begin{array}{l} Equation of straight line joining A (c + 1, e^{c + 1}) \\ and B (c - 1, e^{c - 1}) is \\ y - e^{c + 1} = \frac{e^{c + 1} - e^{c - 1}}{2} (x - c - 1) - - - - - (1) \end{array}$
$Equation of tangent at (c, e^{c}) is$ $y - e^{c} = e^{c} (x - c) - - - - (2)$
Subtracting (1) from (2), we get
$\begin{array}{l} e^{c} (e - 1) = e^{c} [(x - c) - \frac{1}{2} (e - e^{- 1}) (x - c) + \frac{1}{2} (e - e^{- 1})] \\ \Rightarrow \frac{1}{2} (e + e^{- 1}) - 1 = (x - c) [1 - \frac{1}{2} (e - e^{- 1})] \\ \Rightarrow x - c = \frac{e + e^{- 1} - 2}{2 - e + e^{- 1}} < 0 \\ [∴ e + e^{- 1} > 2 and 2 + e^{- 1} - e < 0] \end{array}$
$\begin{array}{l} \Rightarrow x < c \\ Thus the two lines meet to the left of x = c . \end{array}$

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Tangents and normals

Remember concepts with our Masterclasses.

The tangent to the curve y=ex drawn at the point c,ec intersects the line joining the points c−1,ec−1 and c+1,ec+1

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$The tangent to the curve y = e^{x} drawn at the point (c, e^{c}) intersects the$ $line joining the points (c - 1, e^{c - 1}) and (c + 1, e^{c + 1})$