Three normals to the parabola y2=x are drawn through a point (c, 0) then

4a=1⇒a=14Equation of the normaly=mx-2am-am3y=mx−12m−14m3 passing through (c, 0) we get m=0 (or) m=±2c−12For three normals, values of m should be real ∴c−12>0⇒c>12

Three normals to the parabola y2=x are drawn through a point (c, 0) then

4a=1⇒a=14Equation of the normaly=mx-2am-am3y=mx−12m−14m3 passing through (c, 0) we get m=0 (or) m=±2c−12For three normals, values of m should be real ∴c−12>0⇒c>12

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Three normals to the parabola y²=x are drawn through a point (c, 0) then

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$c > \frac{1}{2}$

$c < \frac{1}{2}$

$c = \frac{1}{4}$

$c = \frac{1}{2}$

answer is C.

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Detailed Solution

$4 a = 1 \Rightarrow a = \frac{1}{4}$
Equation of the normal
y=mx-2am-am³
$y = mx - \frac{1}{2} m - \frac{1}{4} m^{3}$ passing through (c, 0) we get m=0 (or) $m = \pm 2 \sqrt{c - \frac{1}{2}}$
For three normals, values of m should be real
$∴ c - \frac{1}{2} > 0 \Rightarrow c > \frac{1}{2}$