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प्रश्न
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).
उत्तर
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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