Q92 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics
The equation of the curve passing through $(1,2)$ and whose tangent at any point $(x, y)$ makes an angle $\tan ^{-1}(2 x+3 y)$ with the $X$-axis is .........
- Correct answerA$6 x+9 y+2=26 e^{3 x-3}$
- B$6 x+9 y-2=26 e^{3 x-3}$
- C$6 x+9 y+2=26 e^{3 x+3}$
- D$6 x+9 y-2=26 e^{3 x+3}$
Solution
$\mathrm{P}=(1,2)$
$$
\begin{aligned}
\text { Inclination }(\theta) & =\tan ^{-1}(2 x+3 y) \\
\tan \theta & =2 x+3 y \\
\therefore \quad \text { Slope }(m) & =\tan \theta
\end{aligned}
$$
$$
\begin{gathered}
\frac{d y}{d x}=2 x+3 y \\
\frac{d y}{d x}-3 y=2 x \\
\therefore P=-3 \text { and } Q=2 x \\
\mathrm{IF}=e^{\int P d x}=e^{\int-3 d x}=e^{-3 x}
\end{gathered}
$$
General soution is,
$$
\begin{aligned}
y(\mathrm{IF}) & =\int Q(\mathrm{IF}) d x+c \\
y\left(e^{-3 x}\right) & =\int 2 x \cdot e^{-3 x} d x+c \\
y\left(e^{-3 x}\right) & =2 x \int e^{-3 x} d x-\int\left(\frac{d}{d x}(2 x) \int e^{-3 x} d x\right) d x+c \\
& =2 x \cdot \frac{e^{-3 x}}{-3}-\int 2 \cdot \frac{e^{-3 x}}{-3} d x+c \\
& =\frac{-2}{3} x e^{-3 x}+\frac{2}{3} \int e^{-3 x} d x+c \\
& =\frac{-2}{3} x e^{-3 x}+\frac{2}{3} \cdot \frac{e^{-3 x}}{-3}+c \\
y e^{-3 x} & =\frac{-2}{3} e^{-3 x}\left(x+\frac{1}{3}\right)+c
\end{aligned}
$$
Multiply by $e^{3 x}$ on both sides
$$
y=-\frac{2}{3}\left(x+\frac{1}{3}\right)+c \cdot e^{3 x}
$$
Since, it is given that Eq. (i) passing through $(1,2)$
$$
\begin{aligned}
2 & =\frac{-2}{3}\left(1+\frac{1}{3}\right)+c \cdot e^{3(1)} \\
2 & =\frac{-8}{9}+c \cdot e^3 \\
\frac{26}{9} & =c e^3 \\
\Rightarrow \quad c & =\frac{26}{9} e^{-3}
\end{aligned}
$$
Put ' $c$ ' value in Eq. (i)
$$
\begin{aligned}
y & =-\frac{2}{3}\left(x+\frac{1}{3}\right)+\frac{26}{9} e^{-3} \cdot e^{3 x} \\
9 y & =-2(3 x+1)+e^{3 x-3}(26) \\
6 x+9 y+2 & =26 \cdot e^{3 x-3}
\end{aligned}
$$
Hence, option (1) is correct.
Hence, the correct answer is option: 'A'.