If the parabolas y² = 4b(x - c) and y² = 8ax have a common normal, then which one of the following i

Q: Wait, why is the x-axis always a common normal?

Great observation! The x-axis is the axis of symmetry for both parabolas. By definition, the normal at the vertex of a parabola is its axis of symmetry. Since both parabolas share the same axis of symmetry, the x-axis is a common normal to both.

Q: Why did we equate the y-intercepts of the two normal lines?

For two lines to be identical, they must have the same slope and the same y-intercept. Since we already assumed they have the same slope $m$, equating the y-intercepts is the necessary and sufficient condition for the two lines to be the same.

Q: Why is $m^2 > 0$ the condition for a non-zero slope?

We are looking for a *real* common normal. If $m^2$ were negative, the slope $m$ would be imaginary, which doesn't exist in the real coordinate plane. Thus, for a valid, non-trivial normal, we must have $m^2 > 0$.

Question

If the parabolas $y^2 = 4b(x - c)$ and $y^2 = 8ax$ have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c)

Options:

(A) (1, 1, 0)

(B) (\frac{1}{2}, 2, 3)

(C) (\frac{1}{2}, 2, 0)

(D) (1, 1, 3)

Sigma Percentile · Accepted Answer

### The Geometry of the Common Normal

Imagine you are standing in a coordinate plane, looking at two parabolas. The first, $y^2 = 8ax$, is a classic, right-opening parabola with its vertex anchored at the origin.

The second, $y^2 = 4b(x - c)$, is its sibling, shifted along the x-axis by a distance $c$. Our mission is to find a line that acts as a normal to both of these curves simultaneously. This is not just an algebraic exercise; it is a search for a geometric bridge between two shapes.

### The Tool: The Slope Form of the Normal

To conquer this, we need the right tool. For any parabola in the standard form $y^2 = 4Ax$, the equation of a normal with slope $m$ is given by the elegant formula:

$$y = mx - 2Am - Am^3$$

For our first parabola, $y^2 = 8ax$, we compare it to $y^2 = 4Ax$ and find that $4A = 8a$, which means $A = 2a$. Substituting this into our normal equation, we get the first normal line:

$$y = mx - 4am - 2am^3 \quad 	ext{--- (i)}$$

### The Shifted Reality

Now, consider the second parabola, $y^2 = 4b(x - c)$. The shift by $c$ means we replace $x$ with $(x - c)$ in our standard normal equation. Here, our $A$ is simply $b$.

So, the normal equation becomes $y = m(x - c) - 2bm - bm^3$. Expanding this, we get:

$$y = mx - mc - 2bm - bm^3 \quad 	ext{--- (ii)}$$

### The Algebraic Dance

Since we are looking for a common normal, Equation (i) and Equation (ii) must represent the same line. They already share the same slope $m$, so for them to be identical, their y-intercepts must be equal.

Setting the intercepts equal, we have:

$$-4am - 2am^3 = -mc - 2bm - bm^3$$

Bringing all terms to one side to see the structure emerge:

$$mc + 2bm - 4am + bm^3 - 2am^3 = 0$$

Factoring out $m$ and $m^3$, we obtain:

$$m(c + 2b - 4a) + m^3(b - 2a) = 0$$

### The Condition for Existence

We can factor out an $m$ from the entire expression:

$$m[ (c + 2b - 4a) + m^2(b - 2a) ] = 0$$

This gives us two cases. The first, $m = 0$, corresponds to the x-axis, which is the axis of symmetry and thus a common normal.

Assuming $m 
eq 0$, we have $(c + 2b - 4a) + m^2(b - 2a) = 0$. Solving for $m^2$, we find:

$$m^2 = \frac{4a - 2b - c}{b - 2a}$$

With a little algebraic manipulation, this simplifies to:

$$m^2 = \frac{c}{2a - b} - 2$$

For a real normal to exist, we must have $m^2 > 0$, which leads us to the **crucial condition**:

$$\frac{c}{2a - b} > 2$$

### The Final Verification

Now, we test our options. For the triad $(a, b, c) = (1, 1, 3)$, we calculate:

$$\frac{3}{2(1) - 1} = 3$$

Since $3 > 2$, the condition is satisfied! We have found our valid choice. This journey through the algebra reveals the underlying geometric constraints that allow these two parabolas to share a common normal.

Animated Solution for Mathematics - Conic Sections: If the parabolas $y^{2} = 4 b (x - c)$ and $y^{2} = 8 a x$ have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c)

Select Answer:

Visualized Solution

Identifying the Parabolas

Normal to Standard Parabola

Normal to $y^{2} = 8 a x$

Normal to $y^{2} = 4 b (x - c)$

Equating the Normals

Rearranging the Equation

The Trivial Case ( $m = 0$ )

Solving for Non-Zero Slope

Simplifying the Expression for $m^{2}$

The Condition for Real Normals

Testing the Options

The Sigma Insight: Equation of Tangent and Normal

The Geometry of the Common Normal

The Tool

The Shifted Reality

The Algebraic Dance

The Condition for Existence

The Final Verification

Similar Questions

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 $α$ .

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 :

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

The tangent $P T$ and the normal $P N$ to the parabola $y^{2} = 4 a x$ at a point $P$ on it meet its axis at points $T$ and $N$ , respectively. The locus of the centroid of the triangle $P T N$ is a parabola whose

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

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

Comprehension Passage

The value of $r$ is

If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

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.

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is