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The normal to a curve at $P(x, y)$ meets the $x$-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is a
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ellipse
ellipse
Equation of normal is $Y-y=-\frac{d x}{d y}(X-x)$
$\begin{aligned}
& \Rightarrow G \equiv\left(x+y \frac{d y}{d x}, 0\right) \\
& \left|x+y \frac{d y}{d x}\right|=|2 x| \\
& \Rightarrow y \frac{d y}{d x}=x \text { or } y \frac{d y}{d x}=-3 x \\
& y d y=x d x \text { or } y d y=-3 x d x \\
& \frac{y^2}{2}=\frac{x^2}{2}+c \text { or } \frac{y^2}{2}=-\frac{3 x^2}{2}+c \\
& x^2-y^2=-2 c \text { or } 3 x^2+y^2=2 c.
\end{aligned}$
$\begin{aligned}
& \Rightarrow G \equiv\left(x+y \frac{d y}{d x}, 0\right) \\
& \left|x+y \frac{d y}{d x}\right|=|2 x| \\
& \Rightarrow y \frac{d y}{d x}=x \text { or } y \frac{d y}{d x}=-3 x \\
& y d y=x d x \text { or } y d y=-3 x d x \\
& \frac{y^2}{2}=\frac{x^2}{2}+c \text { or } \frac{y^2}{2}=-\frac{3 x^2}{2}+c \\
& x^2-y^2=-2 c \text { or } 3 x^2+y^2=2 c.
\end{aligned}$
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