JEE Advanced · Mathematics · 25. AOD

Tangent is drawn at any point $P$ of a curve which passes through $(1,1)$ cutting $X$-axis and $Y$-axis at $A$ and $B$, respectively. If $A P: B P=3: 1$, then

A
differential equation of the curve is $3 x \frac{d y}{d x}+y=0$
B
differential equation of the curve is $3 x \frac{d y}{d x}-y=0$
C
curve is passing through $\left(\frac{1}{8}, 2\right)$
D
normal at $(1,1)$ is $x+3 y=4$

Verified Solution

Answer & Solution

Correct Answer

(C)
curve is passing through $\left(\frac{1}{8}, 2\right)$

Step-by-step Solution

Detailed explanation

Since, $\quad A P: B P=3: 1$
where, equation of tangent is $y-y_1=f^{\prime}(x)\left(x-x_1\right)$
\[
\begin{aligned}
\therefore & A\left(x_1-\frac{y_1}{f^{\prime}\left(x_1\right)}, 0\right) \text { and } B\left(0, y_1-x_1 f^{\prime}(x)\right) \\
& A P: B P=3: 1 \Rightarrow \frac{d y}{d x}=\frac{y}{-3 x} \\
\therefore \quad & 3 x \frac{d y}{d x}+y=0 \quad \text { or } \quad \frac{d y}{y}=-\frac{1}{3 x} d x,
\end{aligned}
\]
On integrating both sides, we get
\[
\begin{array}{lrlr}
\Rightarrow & x y^3 & =1 \\
& \therefore \text { At } & x & =\frac{1}{8} \quad \text { and } \quad y=2
\end{array}
\]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Tangent is drawn at any point \(P\) of a curve which passes through \((1,1)\) cutting \(X\)-axis and \(Y\)-axis at \(A\) and \(B\), respectively. If \(A P: B P=3: 1\), then

Answer & Solution

Step-by-step Solution

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