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I am self-educating myself. I have no tutors. I have joined a test series. It is a question which came in the test. I could not solve it in the test and I cannot solve it now also. So I seek your help.

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    Happy solving in Fiitjee AITS!! – 2017-01-04
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    @Rohan can you solve? – 2017-01-04
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    Maybe the most suitable self-study source of calculus is Howard Anton's. You may get one. – 2017-01-04

1 Answers 1

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Let the given curve be $y = f (x)$. Suppose $P(x, y)$ be a point on the curve.

Equation of the tangent to the curve at $P$ is: $$Y-y = \frac {dy}{dx}(X-x) $$, where $(X, Y)$ is arbitrary point on the tangent.

Putting $Y = 0$, we get the intercept cut off by the tangent on x-axis = $$x-y\frac {dx}{dy} $$ Now given that $$x-y\frac {dx}{dy}=y \Rightarrow \frac {dy}{dx} =\frac {y}{x-y} $$ The equation of the curve satisfying this DE is given by $$yc =e^{-\frac {x}{y}}$$ where $c $ is some constant. Hope you can take it from here.

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    How did you know that it is from FIITJEE AITS? – 2017-01-04