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Question
The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).
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Solution
Let P(x, y) be any point on the curve. Then slope of the tangent at P is \[\frac{dy}{dx}\]
It is given that the slope of the tangent at P(x,y) is equal to the ordinate i.e y.
Therefore \[\frac{dy}{dx}\] = y
\[\Rightarrow \frac{1}{y}dy = dx\]
\[ \Rightarrow \log y = x + \log C\]
\[ \Rightarrow log y = \log e^x + log C\]
\[ \Rightarrow y = C e^x \]
Since, the curve passes through (1,1). Therefore, x=1 and y=1 .
Putting these values in equation obtained above we get,
\[1 = C e^1 \]
\[ \Rightarrow C = \frac{1}{e}\]
putting these values in the equation we get,
\[y = e^{x - 1} \]
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