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JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2006 · Mathematics · Application of Derivatives

Tangent is drawn at any point P of a curve which passes through...

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

Options

Adifferential equation of the curve is $3 x \frac{d y}{d x}+y=0$
Bdifferential equation of the curve is $3 x \frac{d y}{d x}-y=0$
Ccurve is passing through $\left(\frac{1}{8}, 2\right)$
Dnormal at $(1,1)$ is $x+3 y=4$

Correct answer

A. differential equation of the curve is $3 x \frac{d y}{d x}+y=0$

Step-by-step solution

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{\text{ and } } \quad y=2 \end{array}

$

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