Let A and B be two distinct points on the parabola y² = 4x. If the axis of the parabola touches a ci

Q: Wait, why do we use parametric coordinates $(t^2, 2t)$ instead of $(x, y)$?

Great question! While you *could* use $(x_1, y_1)$ and $(x_2, y_2)$, you would quickly find yourself drowning in a sea of variables and constraints like $y_1^2 = 4x_1$. Parametric coordinates are a powerful tool in JEE because they automatically satisfy the parabola's equation, reducing the number of variables and making algebraic manipulation significantly cleaner.

Q: Why is the y-coordinate of the center $t_1 + t_2$?

The center of the circle is the midpoint of the diameter $AB$. The midpoint formula for the y-coordinate is $\frac{y_A + y_B}{2}$. Substituting $y_A = 2t_1$ and $y_B = 2t_2$, we get $\frac{2t_1 + 2t_2}{2}$, which simplifies beautifully to $t_1 + t_2$. It's all about simplifying early!

Q: Why does the circle touching the x-axis mean the radius is $|y_c|$?

Think about the geometry. The distance from any point $(x, y)$ to the x-axis is simply the absolute value of its y-coordinate, $|y|$. If the circle touches the x-axis, the perpendicular distance from the center to the axis is the radius. Therefore, the radius must be the absolute value of the center's y-coordinate.

Question

Let $A$ and $B$ be two distinct points on the parabola $y^2 = 4x$. If the axis of the parabola touches a circle of radius $r$ having $AB$ as its diameter, then the slope of the line joining $A$ and $B$ can be

Options:

(A) $\frac{1}{r}$

(B) $-\frac{1}{r}$

(C) $\frac{2}{r}$

(D) $-\frac{2}{r}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with the parabola $y^2 = 4x$. This is our canvas, a standard rightward-opening parabola where the axis of symmetry is the x-axis, defined by the equation $y = 0$.

### The Power of Parametric Coordinates

We are given two distinct points, $A$ and $B$, on this parabola. To avoid the algebraic burden of Cartesian coordinates, we utilize parametric coordinates.

For a parabola $y^2 = 4ax$ with $a=1$, any point can be represented as $(t^2, 2t)$. Thus, we define our points as $A = (t_1^2, 2t_1)$ and $B = (t_2^2, 2t_2)$.

### The Circle's Dance

Consider a circle with $AB$ as its diameter. The center of this circle, $C$, is the midpoint of the segment $AB$. Using the midpoint formula, the coordinates of $C$ are:

$$C = \left( \frac{t_1^2 + t_2^2}{2}, \frac{2t_1 + 2t_2}{2} ight) = \left( \frac{t_1^2 + t_2^2}{2}, t_1 + t_2 ight)$$

We are told that this circle touches the x-axis. Geometrically, this means the perpendicular distance from the center $C$ to the x-axis is exactly equal to the radius $r$.

The distance from a point $(x, y)$ to the x-axis is $|y|$. Therefore, we have the condition:

$$|t_1 + t_2| = r$$

This implies that $t_1 + t_2 = \pm r$. This result serves as the bridge between our geometric setup and our final answer.

### The Algebraic Symphony

Next, we determine the slope $m$ of the line joining $A$ and $B$. Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$, we substitute our parametric coordinates:

$$m = \frac{2t_2 - 2t_1}{t_2^2 - t_1^2}$$

The denominator is a difference of squares, which factors as $(t_2 - t_1)(t_2 + t_1)$. Since $A$ and $B$ are distinct points, $t_1 
eq t_2$, allowing us to simplify:

$$m = \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$$

### Final Calculation

Substituting the tangency condition $t_1 + t_2 = \pm r$ into our slope equation, we obtain:

$$m = \frac{2}{\pm r} = \pm \frac{2}{r}$$

The possible values for the slope are **$\frac{2}{r}$** and **$-\frac{2}{r}$**. We have successfully navigated the geometry and simplified the algebra to reach this elegant conclusion.

Animated Solution for Mathematics - Conic Sections: Let $A$ and $B$ be two distinct points on the parabola $y^{2} = 4 x$ . If the axis of the parabola touches a circle of radius $r$ having $A B$ as its diameter, then the slope of the line joining $A$ and $B$ can be

Select Answer:

Visualized Solution

Visualizing the Parabola

Points on the Parabola

Parametric Coordinates

The Circle and its Diameter

Coordinates of the Center

The Tangency Condition

Distance from Center to Axis

Equating Radius and y-coordinate

The Goal: Slope of $A B$

Applying the Slope Formula

Simplifying the Slope Expression

Final Substitution

Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Power of Parametric Coordinates

The Circle's Dance

The Algebraic Symphony

Final Calculation

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