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If the length of the sub tangent at any point $p(x, y)$ on a curve $f(x, y)=0$ is $x+7 y^2$, then $f(x, y)=$
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Verified Answer
The correct answer is:
$7 y^2+c y-x$
(c) We know that,
Length of subtangent $=\frac{y}{\frac{d y}{d x}} \Rightarrow x+7 y^2=\frac{y}{\frac{d y}{d x}}$
$$
\begin{array}{ll}
\Rightarrow & \frac{d y}{d x}=\frac{y}{x+7 y^2} \Rightarrow \frac{d x}{d y}=\frac{x+7 y^2}{y} \\
\Rightarrow & \frac{d x}{d y}-\frac{1}{y} x=7 y \quad \therefore \mathrm{IF}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}
\end{array}
$$
So, solution is given by
$$
\begin{aligned}
x \cdot \frac{1}{y} & =\int 7 y \cdot \frac{1}{y} d y+C \Rightarrow \frac{x}{y}=7 y+C \\
x & =7 y^2+c y \Rightarrow 7 y^2+c y-x=0 \\
\therefore \quad f(x, y) & : 7 y^2+c y-x=0
\end{aligned}
$$
Length of subtangent $=\frac{y}{\frac{d y}{d x}} \Rightarrow x+7 y^2=\frac{y}{\frac{d y}{d x}}$
$$
\begin{array}{ll}
\Rightarrow & \frac{d y}{d x}=\frac{y}{x+7 y^2} \Rightarrow \frac{d x}{d y}=\frac{x+7 y^2}{y} \\
\Rightarrow & \frac{d x}{d y}-\frac{1}{y} x=7 y \quad \therefore \mathrm{IF}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}
\end{array}
$$
So, solution is given by
$$
\begin{aligned}
x \cdot \frac{1}{y} & =\int 7 y \cdot \frac{1}{y} d y+C \Rightarrow \frac{x}{y}=7 y+C \\
x & =7 y^2+c y \Rightarrow 7 y^2+c y-x=0 \\
\therefore \quad f(x, y) & : 7 y^2+c y-x=0
\end{aligned}
$$
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