Differential Equations Ques 62
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 $3 x \frac{d y}{d x}-y=0$
(c) curve is passing through $(\frac{1}{8}, 2)$
(d) normal at $(1,1)$ is $x+3 y=4$.
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Answer:
Correct Answer: 62.(a, c)
Solution:
- Since, $B P: A P=3: 1$. Then, equation of tangent is
$ Y-y=f^{\prime}(x)(X-x) $
The intercept on the coordinate axes are
$ A (x-\frac{y}{f^{\prime}(x)}, 0) $
and
$ B\left[0, y-x f^{\prime}(x)\right] $
Since, $P$ is internally intercepts a line $A B$,
$\therefore \quad x=\frac{3 (x-\frac{y}{f^{\prime}(x)})+1 \times 0}{3+1} $
$\Rightarrow \frac{d y}{d x}=\frac{y}{-3 x} \Rightarrow \frac{d y}{y}=-\frac{1}{3 x} d x$
On integrating both sides, we get
$ x y^{3}=c $
Since, curve passes through $(1,1)$, then $c=1$.
$\therefore \quad x y^{3}=1$
At $\quad x=\frac{1}{8} \Rightarrow y=2$
Hence, (a) and (c) are correct answers.
BETA
