Three normal to the parabola $y^2=x$ are drawn through the point .

Question

Three normal to the parabola $y^2=x$ are drawn through the point $(c,0)$ then $$\textrm {a}. c=\dfrac {1}{4}$$ $$\textrm {b}. c=1$$ $$\textrm {c}. c>\dfrac {1}{2}$$ $$\textrm {d}. c=\dfrac {1}{2}$$

My Attempt: Comparing $y^2=x$ with $y^2=4ax$ we get, $$4a=1$$ $$a=\dfrac {1}{4}$$ The equation of normal to the parabola $y^2=4ax$ with slope $m$ is $$y=mx-2am-am^3$$ $$y=mx-2\dfrac {1}{4} m -\dfrac {1}{4} m^3$$ This equation passes through $(c,0)$. So $$0=mc-\dfrac {m}{2}-\dfrac {m^3}{4}$$

Answer 1

Consider this: If we draw the three normals at $(a^2, a)$, $(a^2, -a)$ and $(0, 0)$ for any positive number $a$, then they intersect at some $(c, 0)$. Note that we must have $c>a$. There is only one option that allows for this. That's the answer. (If you think that's cheap, then you see why the choices in multiple choice questions really need some thought.)

Actual answer: Take some positive number $a$, and consider the point $(a^2, a)$. The tangent to the parabola at that point has slope $\frac{1}{2a}$. This means that the normal has slope $-2a$.

So the normal is a line that goes through the point $(a^2, a)$ and has slope $-2a$, that means it intersects the $x$-axis at $(a^2 + \frac12, 0)$. Thus we have $c = a^2 + \frac12$, so clearly (c) is the correct option.

Answer 2

Hint:

Using https://www.askiitians.com/iit-jee-coordinate-geometry/normal-to-a-parabola.aspx

the equation of normal $y=mx-2am-am^3$ where $4a=1$

Now this normal has to pass through $(c,0)$

If $m\ne0$ $$am^2+2a-c=0$$

Now this equation needs to have two non-zero district roots

Answer 3

Evolute of parabola y^2= a x

A comprehensive picture is obtained by sketching its above evolute. Normals at vertex cut x-axis at sphere center locally:

$$ c\ge \frac24 =\frac 12 $$

Normals to parabola intersect x-axis at or beyond this center of curvature point. Option 3.