Question

At a point $P(x,y)$ on a curve $x=f(y)$, the x-intercept of the tangent is always equal to the y-coordinate of the point of contact, then $f(y)=$

• A. $e^{cy^{2}}$
• B. $y\log\left(\frac{c}{y}\right)$
• C. $cy^{2}$
• D. $\sin(c+y)$

Hint:

The formula for the x-intercept of a tangent line is $X = x - y\frac{dx}{dy}$, while the y-intercept is $Y = y - x\frac{dy}{dx}$.

Answer 1

The Correct Option is B

Step 1: Write the tangent line.
At $P(x,y)$ on $x=f(y)$, the tangent is $Y-y=\dfrac{dy}{dx}(X-x)$. We find its x-intercept by setting $Y=0$.
Step 2: Find the x-intercept.
Setting $Y=0$: $-y=\dfrac{dy}{dx}(X-x)$, so $X=x-y\dfrac{dx}{dy}$.
Step 3: Use the given condition.
The x-intercept equals the y-coordinate of $P$, so $x-y\dfrac{dx}{dy}=y$.
Step 4: Form a linear equation in $x$.
Rearrange to $y\dfrac{dx}{dy}=x-y$, then divide by $y$: \[ \frac{dx}{dy}-\frac{1}{y}\,x=-1. \]
Step 5: Solve with an integrating factor.
The integrating factor is $e^{\int -\frac{1}{y}dy}=\dfrac{1}{y}$. Then $\dfrac{x}{y}=\int -\dfrac{1}{y}\,dy=-\log y+c$.
Step 6: Write $x=f(y)$.
\[ x=y(c-\log y)=y\log\!\left(\frac{c}{y}\right). \] \[ \boxed{f(y)=y\log\!\left(\dfrac{c}{y}\right)} \]