SATHEE: Differential Equations 3 Question 7

Differential Equations 3 Question 7

7.

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 $B P : A P = 3 : 1$ , then

(2006, 3M)

(a) differential equation of the curve is $3 x \frac{d y}{d x} + y = 0$

(b) differential equation of the curve is $3 x \frac{d y}{d x} - y = 0$

(d) normal at $(1, 1)$ is $x + 3 y = 4$ .

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Answer:

Correct Answer: 7. (a, c)

Solution:

Since, $B P : A P = 3 : 1$ . Then, equation of tangent is

$Y - y = f^{'} (x) (X - x)$

The intercept on the coordinate axes are

$A (x - \frac{y}{f^{'} (x)}, 0)$

and

$B [0, y - x f^{'} (x)]$

Since, $P$ is internally intercepts a line $A B$ ,

$∴ x = \frac{3 (x - \frac{y}{f^{'} (x)}) + 1 \times 0}{3 + 1}$

alt text

$\Rightarrow \frac{d y}{d x} = \frac{y}{- 3 x} \Rightarrow \frac{d y}{y} = - \frac{1}{3 x} d x$

On integrating both sides, we get

$x y^{3} = c$

Since, curve passes through $(1, 1)$ , then $c = 1$ .

$∴ x y^{3} = 1$

At $x = \frac{1}{8} \Rightarrow y = 2$

Hence, (a) and (c) are correct answers.

Differential Equations 3 Question 7

7.

Answer:

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Differential Equations 3 Question 7

7.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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