Three normals are drawn from the point (c, 0) to the curve y² = x. Show that c must be greater than

Q: Wait, why do we use $y^2 = 4ax$ when the problem gives $y^2 = x$?

Great observation! The form $y^2 = 4ax$ is the 'universal' language for parabolas. By comparing $y^2 = x$ to $y^2 = 4ax$, we identify that $4a = 1$, so $a = 1/4$. This allows us to use the standard, pre-derived formula for the normal, saving us from performing tedious calculus every single time.

Q: Why does $c > 1/2$ guarantee three normals?

Think of the cubic equation $m(m^2 - 4(c - 1/2)) = 0$. One root is $m=0$ (the x-axis). For the other two to exist and be distinct, the term inside the parenthesis must yield two real, non-zero values. If $c = 1/2$, the roots are $0, 0, 0$ (coincident). If $c 1/2$ is the 'magic threshold' for the geometry to allow three distinct normal lines.

Q: Can I just use the slope formula $m = -dy/dx$?

You absolutely can! If you differentiate $y^2 = x$ to get $2y(dy/dx) = 1$, you find the slope of the tangent is $1/(2y)$. The normal slope is then $-2y$. By setting the line equation $Y - 0 = -2y(X - x)$ to pass through $(c, 0)$, you will arrive at the same cubic equation. The slope-form method is just a faster shortcut!

Q: Why is the product of roots $-4(c - 1/2)$ and not just $4(c - 1/2)$?

Look closely at the quadratic equation $m^2 - 4(c - 1/2) = 0$. In the standard form $Am^2 + Bm + C = 0$, our $C$ is $-4(c - 1/2)$. The product of roots is $C/A$. Since $A=1$, the product is indeed $-4(c - 1/2)$. Always watch your signs—they are where the most brilliant students often trip!

Q: Is the x-axis always a normal?

Yes, for any parabola symmetric about the x-axis, the x-axis itself is a normal at the vertex. Because the tangent at the vertex $(0,0)$ is the y-axis (vertical), any line perpendicular to it—which is the x-axis—must be a normal. It is a beautiful geometric property that never changes!

Question

Three normals are drawn from the point $(c, 0)$ to the curve $y^2 = x$. Show that $c$ must be greater than $1/2$. One normal is always the x-axis. Find $c$ for which the other two normals are perpendicular to each other.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the vertex of a perfectly smooth, reflective parabolic mirror defined by the equation $y^2 = x$. You are holding a laser pointer at a point $(c, 0)$ on the axis of symmetry.

You want to fire beams that hit the parabola such that they strike it 'normally'—meaning they hit the surface at a perfect 90-degree angle. We aim to determine how many such beams can be fired and the condition under which two of those beams are perpendicular to each other.

### The Universal Language of Parabolas

To solve this, we first need to speak the language of parabolas. While our curve is $y^2 = x$, the standard form $y^2 = 4ax$ is our most powerful tool.

By comparing the two, we immediately see that $4a = 1$, which gives us the focal parameter $a = 1/4$. This $a$ is the heartbeat of the parabola; it dictates how 'wide' or 'narrow' the curve opens.

Now, we invoke the general equation of a normal to a parabola in slope form: $y = mx - 2am - am^3$. This equation is a masterpiece of coordinate geometry.

Substituting our value of $a = 1/4$, the equation simplifies to:

$$y = mx - \frac{m}{2} - \frac{m^3}{4}$$

### The Point of Intersection

We are firing our normals from the point $(c, 0)$. This means that for any normal we draw, the coordinates $(c, 0)$ must satisfy the equation of that line.

Plugging these into our equation, we get:

$$0 = mc - \frac{m}{2} - \frac{m^3}{4}$$

If we factor out $m$, we are left with a beautiful cubic structure:

$$m \left( c - \frac{1}{2} - \frac{m^2}{4} \right) = 0$$

This equation is the key to the entire mystery. It tells us that there are three possible slopes for our normals. The first solution is $m = 0$, which corresponds to the x-axis itself, confirming that the axis of symmetry is always a normal to the parabola.

### The Threshold of Existence

But what about the other two normals? They come from the quadratic part: $c - 1/2 - m^2/4 = 0$.

Rearranging this, we find $m^2 = 4(c - 1/2)$. For these two normals to actually exist as real lines, $m^2$ must be positive.

This forces the condition $c - 1/2 > 0$, or simply $c > 1/2$. If $c$ were less than $1/2$, the normals would vanish into the realm of imaginary numbers. We have just proven that to see three distinct normals, our point must be far enough away from the vertex.

### The Grand Finale: Perpendicularity

Finally, we reach the most thrilling part of our journey. We want the two non-axial normals to be perpendicular.

If their slopes are $m_1$ and $m_2$, the condition for perpendicularity is $m_1 m_2 = -1$. From our quadratic equation $m^2 - 4(c - 1/2) = 0$, we can identify the product of the roots.

In a quadratic $m^2 + 0m + C = 0$, the product of the roots is simply the constant term. Here, that constant is $-4(c - 1/2)$. Setting this equal to $-1$, we get:

$$-4\left(c - \frac{1}{2}\right) = -1$$

Solving this simple linear equation:

$$4c - 2 = 1 \Rightarrow 4c = 3 \Rightarrow c = \frac{3}{4}$$

And there it is! When **$c = 3/4$**, the two normals are perfectly perpendicular. You have navigated the cubic landscape, respected the geometric constraints, and arrived at the precise coordinate where the physics of the parabola aligns in perfect harmony.

Animated Solution for Mathematics - Conic Sections: Three normals are drawn from the point $(c, 0)$ to the curve $y^{2} = x$ . Show that $c$ must be greater than $1/2$ . One normal is always the x-axis. Find $c$ for which the other two normals are perpendicular to each other.

Enter Numerical Value:

Visualized Solution

Visualizing the Parabola and Point $(c, 0)$

General Equation of a Normal

Finding the value of $a$

Substituting $a$ into the Normal Equation

Passing the Normal through $(c, 0)$

Factorizing the Equation

The First Normal (The x-axis)

Condition for the Other Two Normals

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

The Perpendicularity Condition

Using the Product of Roots

Solving for $c$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Universal Language of Parabolas

The Point of Intersection

The Threshold of Existence

The Grand Finale

Similar Questions

If the three normals drawn to the parabola, $y^{2} = 2 x$ pass through the point $(a, 0) a \neq = 0$ , then 'a' must be greater than :

Suppose that the normals drawn at three different points on the parabola $y^{2} = 4 x$ pass through the point $(h, k)$ . Show that $h > 2$ .

Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

Match the following : $(3, 0)$ is the point from which three normals are drawn to the parabola $y^{2} = 4 x$ which meet the parabola in the points $P, Q$ and $R$ . Then

List-I

List-II

What normal to the curve $y = x^{2}$ forms the shortest chord?

Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ intersect a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(- \frac{1}{3 2}, 0)$ and $(0, β)$ , then $β$ is equal to :

Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(\frac{1}{3 2}, 0)$ and $(0, β)$ , then $β$ is equal to :

The number of values of $c$ such that the straight line $y = 4 x + c$ touches the curve $(x^{2} /4) + y^{2} = 1$ is

The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^{2} + y^{2} = 4$ at $(1, rac{\sqrt{3}}{1})$ is .........

Let $P (a sec θ, b tan θ)$ and $Q (a sec ϕ, b tan ϕ)$ , where $θ + ϕ = π /2$ , be two points on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$ , then $k$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Parabola and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (c,0)