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$\mathrm{A}(-4,0)$ and $\mathrm{B}(4,0)$ are two fixed points. $\mathrm{C}$ and $\mathrm{D}$ are two points on $\mathrm{Y}$-axis such that $\mathrm{CD}=4$ and $\mathrm{C}$ is a point below $\mathrm{D}$. Then the locus of the point of intersection of the lines $\mathrm{AC}$ and $\mathrm{BD}$ is
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Verified Answer
The correct answer is:
$x^2+2 x y-16=0$
Let- $\mathrm{C}\left(0_1 \mathrm{y}_1\right)$ and $\mathrm{D}\left(0_1 \mathrm{y}_2\right)$
$\therefore$ Equation of $\mathrm{AC}$
$$
\begin{aligned}
& \mathrm{y}-\mathrm{y}_1=\frac{0-\mathrm{y}_1}{-4-0}(\mathrm{x}-0) \\
& \mathrm{y}-\mathrm{y}_1=\frac{\mathrm{y}_1 \mathrm{x}}{4} \Rightarrow \mathrm{y}_1=\frac{4 \mathrm{y}}{\mathrm{x}+4}
\end{aligned}
$$
Equation of BD
$$
\begin{aligned}
& y-y_2=\frac{0-y_2}{4-0}(x-0) \\
& y-y_2=\frac{-y_2 x}{4} \Rightarrow y_2=\frac{4 y}{4-x} \ldots \ldots . .
\end{aligned}
$$
Given that $\mathrm{y}_2-\mathrm{y}_1=4$
$$
\begin{aligned}
& \therefore \frac{4 y}{4-x}-\frac{4 y}{x+4}=4 \\
& \Rightarrow \frac{4 y(x+4-4+x)}{16-x^2}=4 \\
& \Rightarrow y(2 x)=16-x^2 \\
& \Rightarrow x^2+2 x y-16=0
\end{aligned}
$$
$\therefore$ Equation of $\mathrm{AC}$
$$
\begin{aligned}
& \mathrm{y}-\mathrm{y}_1=\frac{0-\mathrm{y}_1}{-4-0}(\mathrm{x}-0) \\
& \mathrm{y}-\mathrm{y}_1=\frac{\mathrm{y}_1 \mathrm{x}}{4} \Rightarrow \mathrm{y}_1=\frac{4 \mathrm{y}}{\mathrm{x}+4}
\end{aligned}
$$
Equation of BD
$$
\begin{aligned}
& y-y_2=\frac{0-y_2}{4-0}(x-0) \\
& y-y_2=\frac{-y_2 x}{4} \Rightarrow y_2=\frac{4 y}{4-x} \ldots \ldots . .
\end{aligned}
$$
Given that $\mathrm{y}_2-\mathrm{y}_1=4$
$$
\begin{aligned}
& \therefore \frac{4 y}{4-x}-\frac{4 y}{x+4}=4 \\
& \Rightarrow \frac{4 y(x+4-4+x)}{16-x^2}=4 \\
& \Rightarrow y(2 x)=16-x^2 \\
& \Rightarrow x^2+2 x y-16=0
\end{aligned}
$$
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