A is a point on the parabola y² = 4ax. The normal at A cuts the parabola again at point B. If AB sub

Q: Why can't I just find the coordinates of A and B and use the slope formula?

You absolutely could, but it's a trap! You would end up solving a quartic equation or dealing with complex parametric coordinates $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$. Homogenization bypasses the need to know the specific coordinates of $A$ and $B$ entirely, saving you precious time and reducing the chance of calculation errors.

Q: Wait, why do we reject $m=0$ at the end?

Great catch! If $m=0$, the normal is the $x$-axis. The $x$-axis is the axis of the parabola. It doesn't intersect the parabola at a point $B$ to form a triangle with the vertex. It is a degenerate case that doesn't satisfy the geometric condition of forming a right angle at the vertex.

Q: Is the condition $A+C=0$ for perpendicular lines always true?

Yes, for any second-degree homogeneous equation $Ax^2 + Bxy + Cy^2 = 0$, the lines are perpendicular if and only if the sum of the coefficients of $x^2$ and $y^2$ is zero. It is a beautiful, time-saving property derived from the fact that the product of slopes is $-1$.

Question

$A$ is a point on the parabola $y^2 = 4ax$. 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 $AB$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the classic parabola $y^2 = 4ax$. You pick a point $A$ on this curve and draw a normal line. This line cuts through the parabola again at a point $B$.

The problem provides a specific condition: if you connect the vertex $O(0,0)$ to $A$ and $B$, the angle between these two lines is exactly $90^\circ$. Our goal is to find the slope $m$ of this normal line $AB$.

### The Power of Homogenization

Many students attempt to solve this by finding the coordinates of $A$ and $B$ using parametric forms $(at^2, 2at)$. While valid, this is often a long, winding road filled with algebraic pitfalls.

Instead, we use a technique favored by JEE examiners: **Homogenization**. We aim to find the joint equation of the lines $OA$ and $OB$. Since these lines pass through the origin, their joint equation must be a homogeneous equation of the second degree, taking the form:

$$Ax^2 + Bxy + Cy^2 = 0$$

### The Magic Key

We start with the equation of the normal to the parabola $y^2 = 4ax$ in terms of its slope $m$:

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

To homogenize the parabola's equation, we must express the line equation in a form equal to $1$. We rearrange the normal equation as follows:

$$mx - y = 2am + am^3$$

Dividing both sides by the constant term, we obtain our "magic key":

$$\frac{mx - y}{2am + am^3} = 1$$

### The Transformation

Now, we take the parabola equation $y^2 = 4ax$. The $y^2$ term is already of degree two, but the $4ax$ term is only degree one. We multiply $4ax$ by our magic fraction to "boost" its degree to two:

$$y^2 = 4ax \left( \frac{mx - y}{2am + am^3} \right)$$

Notice how the $a$ in the numerator and the $a$ in the denominator cancel out perfectly. Cross-multiplying gives:

$$y^2(2m + m^3) = 4x(mx - y)$$

Expanding this, we get $y^2(2m + m^3) = 4mx^2 - 4xy$. Rearranging everything to one side, we arrive at the standard form:

$$4mx^2 - 4xy - (2m + m^3)y^2 = 0$$

### The Final Victory

We know that for a pair of lines represented by $Ax^2 + Bxy + Cy^2 = 0$ to be perpendicular, the sum of the coefficients of $x^2$ and $y^2$ must be zero. In our equation, the coefficient of $x^2$ is $4m$, and the coefficient of $y^2$ is $-(2m + m^3)$.

Setting their sum to zero:

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

This simplifies to $2m - m^3 = 0$, or $m(2 - m^2) = 0$. We reject $m=0$ because it represents the axis of the parabola, which does not form a triangle with the origin.

Thus, $m^2 = 2$, and the slope is **$m = \pm \sqrt{2}$**. You have successfully conquered this problem using the elegance of homogenization.

Animated Solution for Mathematics - Conic Sections: $A$ 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 $A B$ .

Visualized Solution

Visualizing the Geometry

Equation of the Normal

The Concept of Homogenization

Preparing the Line for Homogenization

Homogenizing the Parabola

Simplifying the Equation

Expanding the Terms

Standard Quadratic Form

The Perpendicularity Condition

Applying the Condition

Solving for Slope $m$

Conclusion and Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of Homogenization

The Magic Key

The Transformation

The Final Victory

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