A(-4,0) and B(4,0) are two fixed points. C an | vedclass

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(B) Let $C(0, y_1)$ and $D(0, y_2)$ be the points on the $Y$-axis. Since $C$ is below $D$ and $CD=4$,we have $y_2 - y_1 = 4$. The equation of line $AC

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$x^2+2xy-16=0$

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(B) Let $C(0, y_1)$ and $D(0, y_2)$ be the points on the $Y$-axis. Since $C$ is below $D$ and $CD=4$,we have $y_2 - y_1 = 4$. The equation of line $AC$ passing through $A(-4, 0)$ and $C(0, y_1)$ is: $y - 0 = \frac{y_1 - 0}{0 - (-4)}(x - (-4))$ $\Rightarrow y = \frac{y_1}{4}(x + 4)$ $\Rightarrow y_1 = \frac{4y}{x+4}$. The equation of line $BD$ passing through $B(4, 0)$ and $D(0, y_2)$ is: $y - 0 = \frac{y_2 - 0}{0 - 4}(x - 4)$ $\Rightarrow y = \frac{y_2}{-4}(x - 4)$ $\Rightarrow y_2 = \frac{4y}{4-x}$. Given $y_2 - y_1 = 4$,we substitute the expressions for $y_1$ and $y_2$: $\frac{4y}{4-x} - \frac{4y}{x+4} = 4$. Dividing by $4$: $\frac{y}{4-x} - \frac{y}{x+4} = 1$. $\frac{y(x+4) - y(4-x)}{(4-x)(x+4)} = 1$. $\frac{xy + 4y - 4y + xy}{16 - x^2} = 1$. $2xy = 16 - x^2$. $x^2 + 2xy - 16 = 0$.

$A(-4,0)$ and $B(4,0)$ are two fixed points. $C$ and $D$ are two points on the $Y$-axis such that $CD=4$ and $C$ is a point below $D$. Then the locus of the point of intersection of the lines $AC$ and $BD$ is

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