Normals are drawn from the point P with slopes m1, m2, m3 to the parabola y² = 4x. If locus of P wit

Q: Wait, why do we assume $m_1 m_2 = \alpha$ implies $m_3 = -k/\alpha$?

Great question! We use Vieta's formulas for the cubic equation $m^3 + (2-h)m + k = 0$. The product of the roots $m_1 m_2 m_3$ is equal to the negative of the constant term, which is $-k$. Since we are given $m_1 m_2 = \alpha$, we simply substitute this into the product: $\alpha m_3 = -k$. Solving for $m_3$ gives us $m_3 = -k/\alpha$.

Q: Why does the locus of $P$ have to match the original parabola?

The problem explicitly states that the locus of $P$ is a part of the parabola $y^2 = 4x$ itself. This means that the equation we derive for the locus, $y^2 = \alpha^2 x + \alpha^2(\alpha - 2)$, must be identical to $y^2 = 4x$. By comparing the coefficients of $x$ and the constant terms, we can solve for $\alpha$.

Q: Can we use the property of the normal at a point $(at^2, 2at)$ instead?

You absolutely can! The normal at a point $(t^2, 2t)$ on $y^2 = 4x$ is $y = -tx + 2t + t^3$. If this passes through $(h, k)$, then $k = -th + 2t + t^3$, or $t^3 + (2-h)t + k = 0$. This is essentially the same cubic equation, just with $t$ instead of $m$. The slope $m$ is $-1/t$, so the logic remains perfectly consistent.

Question

Normals are drawn from the point $P$ with slopes $m_1, m_2, m_3$ to the parabola $y^2 = 4x$. If locus of $P$ with $m_1m_2 = \alpha$ is a part of the parabola itself then find $\alpha$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the curve of a parabola, $y^2 = 4x$. You are holding a laser, and you want to draw a normal line—a line perpendicular to the tangent at that point.

From a specific point $P(h, k)$ in the plane, you find that you can draw not just one, but three distinct normal lines to this parabola. This is a classic scenario in coordinate geometry.

We are given that the product of the slopes of two of these normals is a constant $\alpha$, and we are told that the locus of $P$ is the parabola itself. Let's unravel this mystery together.

### The Normal Equation in Slope Form

To begin, we need the equation of a normal to the parabola $y^2 = 4ax$ in terms of its slope $m$. The standard form is $y = mx - 2am - am^3$.

For our specific parabola, $y^2 = 4x$, we have $a = 1$. Thus, the equation of any normal with slope $m$ is:

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

This equation is our gateway. It tells us that for any normal passing through a point $(x, y)$, the slope $m$ must satisfy this relationship.

### The Cubic Connection

Since the normals are drawn from the point $P(h, k)$, the coordinates of $P$ must satisfy the normal equation. Substituting $x=h$ and $y=k$, we get $k = mh - 2m - m^3$.

Rearranging this, we form a cubic equation in $m$:

$$m^3 + (2-h)m + k = 0$$

This is a beautiful result. A cubic equation in $m$ means there are three possible slopes for the normals passing through $P$. Let these slopes be $m_1, m_2, m_3$, which are the roots of our cubic equation.

### Vieta's Relations: The Bridge

Now, we invoke the power of Vieta's formulas. For a cubic equation of the form $m^3 + Am + B = 0$, the product of the roots is $-B$.

In our case, the constant term is $k$, so the product of the slopes is:

$$m_1 m_2 m_3 = -k$$

We are given the condition $m_1 m_2 = \alpha$. Substituting this into our product relation, we get $\alpha m_3 = -k$, which leads us to $m_3 = -k/\alpha$.

We have successfully expressed one of the slopes in terms of the coordinates of $P$ and the constant $\alpha$.

### The Locus of P

Since $m_3$ is a root of the cubic equation, it must satisfy it. Let's substitute $m = -k/\alpha$ into $m^3 + (2-h)m + k = 0$:

$$\left(-\frac{k}{\alpha}ight)^3 + (2-h)\left(-\frac{k}{\alpha}ight) + k = 0$$

Expanding this, we get:

$$-\frac{k^3}{\alpha^3} - \frac{k(2-h)}{\alpha} + k = 0$$

Assuming $k 
eq 0$, we can divide by $-k$ to get:

$$\frac{k^2}{\alpha^3} + \frac{2-h}{\alpha} - 1 = 0$$

Multiplying by $\alpha^3$, we obtain $k^2 + \alpha^2(2-h) - \alpha^3 = 0$. Rearranging for $k^2$, we find:

$$k^2 = \alpha^2 h + \alpha^3 - 2\alpha^2$$

Replacing $(h, k)$ with $(x, y)$, the locus is $y^2 = \alpha^2 x + \alpha^2(\alpha - 2)$.

### The Final Reveal

We are told this locus is the parabola $y^2 = 4x$. Comparing the two equations, we match the coefficients:

$$\alpha^2 = 4 \quad 	ext{and} \quad \alpha^2(\alpha - 2) = 0$$

From $\alpha^2 = 4$, we get $\alpha = 2$ or $\alpha = -2$. From $\alpha^2(\alpha - 2) = 0$, we get $\alpha = 0$ or $\alpha = 2$.

The only value that satisfies both is **$\alpha = 2$**. We have arrived at the solution!

Animated Solution for Mathematics - Conic Sections: 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 $α$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Normals

The Normal Equation in Slope Form

Substituting Point $P$

Forming the Cubic Equation

Applying Vieta's Relations

Using the Given Condition

Substituting the Root

Simplifying the Equation

Finding the Locus Equation

Comparing with the Original Parabola

Matching Coefficients

Final Value of $α$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal Equation in Slope Form

The Cubic Connection

Vieta's Relations

The Locus of P

The Final Reveal

Similar Questions

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

$A$ is a point on the parabola $y^{2} = 4 a x$ . The normal at $A$ cuts the parabola again at point $B$ . If $A B$ subtends a right angle at the vertex of the parabola, find the slope of $A B$ .

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

If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

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

If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ , then the value of $r^{2}$ is

Let $L$ be a normal to the parabola $y^{2} = 4 x$ . If $L$ passes through the point $(9, 6)$ , then $L$ is given by

Enter Numerical Value:

Visualized Solution

Visualizing the Normals

The Normal Equation in Slope Form

Substituting Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Forming the Cubic Equation

Applying Vieta's Relations

Using the Given Condition

Substituting the Root

Simplifying the Equation

Finding the Locus Equation

Comparing with the Original Parabola

Matching Coefficients

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal Equation in Slope Form

The Cubic Connection

Vieta's Relations

The Locus of P

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose that the normals drawn at three different points on the parabola y2=4x pass through the point (h,k). Show that h>2.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A is a point on the parabola y2=4ax. The normal at A cuts the parabola again at point B. If AB subtends a right angle at the vertex of the parabola, find the slope of AB.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the three normals drawn to the parabola, y2=2x pass through the point (a,0)a=0, then 'a' must be greater than :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Match the following : (3,0) is the point from which three normals are drawn to the parabola y2=4x which meet the parabola in the points P,Q and R. Then

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two tangents drawn from a point P to the parabola y2=4x are at right angles, then the locus of P is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of r is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If st=1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normals of the parabola y2=4x drawn at the end points of its latus rectum are tangents to the circle (x−3)2+(y+2)2=r2, then the value of r2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let L be a normal to the parabola y2=4x. If L passes through the point (9,6), then L is given by

Substituting Point $P$

Final Value of $α$

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

$A$ is a point on the parabola $y^{2} = 4 a x$ . The normal at $A$ cuts the parabola again at point $B$ . If $A B$ subtends a right angle at the vertex of the parabola, find the slope of $A B$ .

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

If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

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

If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ , then the value of $r^{2}$ is

Let $L$ be a normal to the parabola $y^{2} = 4 x$ . If $L$ passes through the point $(9, 6)$ , then $L$ is given by