The equation of the lines joining the vertex | vedclass

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(D) The equation of the parabola is $y^2 = 6x$.Given the abscissa (x-coordinate) is $24$,we substitute $x = 24$ into the equation:$y^2 = 6(24) = 144$$

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$2y \pm x = 0$ and $x \pm 2y = 0$

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(D) The equation of the parabola is $y^2 = 6x$.Given the abscissa (x-coordinate) is $24$,we substitute $x = 24$ into the equation:$y^2 = 6(24) = 144$$y = \pm 12$.Thus,the points on the parabola are $(24, 12)$ and $(24, -12)$.The vertex of the parabola $y^2 = 6x$ is $(0, 0)$.The equation of the line joining $(0, 0)$ and $(24, 12)$ is $y - 0 = \frac{12 - 0}{24 - 0}(x - 0)$ $\Rightarrow y = \frac{1}{2}x$ $\Rightarrow 2y = x$.The equation of the line joining $(0, 0)$ and $(24, -12)$ is $y - 0 = \frac{-12 - 0}{24 - 0}(x - 0)$ $\Rightarrow y = -\frac{1}{2}x$ $\Rightarrow 2y = -x$.Combining these,we get $2y = \pm x$,which can also be written as $x \pm 2y = 0$.

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

The equation of the lines joining the vertex of the parabola $y^2 = 6x$ to the points on it whose abscissa is $24$ is:

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