Three normals are drawn from the point (c, 0) | vedclass

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(C) The given curve is $y^2 = x$,which is a parabola of the form $y^2 = 4ax$ with $a = \frac{1}{4}$.The equation of a normal to the parabola $y^2 = 4a

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$\frac{3}{4}$

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(C) The given curve is $y^2 = x$,which is a parabola of the form $y^2 = 4ax$ with $a = \frac{1}{4}$.The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$.Substituting $a = \frac{1}{4}$,we get $y = mx - \frac{m}{2} - \frac{m^3}{4}$.Since the normal passes through $(c, 0)$,we have $0 = mc - \frac{m}{2} - \frac{m^3}{4}$.This simplifies to $m(c - \frac{1}{2} - \frac{m^2}{4}) = 0$.Thus,$m = 0$ (which is the $X$-axis) or $m^2 = 4c - 2$.The slopes of the other two normals are $m_1 = \sqrt{4c - 2}$ and $m_2 = -\sqrt{4c - 2}$.Since these two normals are perpendicular,their product must be $-1$,so $m_1 m_2 = -1$.Therefore,$-(\sqrt{4c - 2})^2 = -1$,which implies $4c - 2 = 1$.Solving for $c$,we get $4c = 3$,so $c = \frac{3}{4}$.

$A. \ 2x-5=0$	$I. \ \text{Vertex}$
$B. \ (\frac{3}{2}, -1)$	$II. \ \text{Focus}$
$C. \ y+1=0$	$III. \ \text{Equation of directrix}$
$D. \ (2, -1)$	$IV. \ \text{Equation of the axis}$
	$V. \ \text{Equation of the Latus rectum}$

Three normals are drawn from the point $(c, 0)$ to the curve $y^2=x$. If one of the normals is the $X$-axis,then the value of $c$ for which the other two normals are perpendicular to each other is

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