Let the tangent to the parabola y² = 12x at the point (3, α) be perpendicular to the line 2x + 2y =

Q: Wait, why did we get two values for $\alpha$ (plus and minus 6) but only use one? How do I know which one to pick?

That is a fantastic observation! In algebra, squaring often introduces extraneous solutions. When we solved $\alpha^2 = 36$, we got both $\pm 6$. However, the geometric condition of perpendicularity acts as a filter. By substituting both into the slope condition, you would find that only one value satisfies the specific orientation required by the problem. Always let the geometry guide your algebraic choices.

Q: Is there a shortcut for the normal equation? Do I really need to memorize $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$?

While you can derive the normal using calculus (finding the tangent slope and taking the negative reciprocal), memorizing the standard form for the normal is a massive time-saver in JEE Advanced. It is elegant and symmetric. Think of it as a tool in your belt—the more you use it, the more natural it becomes.

Q: Why is the distance formula modulus important? I keep getting a negative distance.

Distance is a physical quantity, and in our universe, it cannot be negative! The modulus $|Ax_0 + By_0 + C|$ is there to ensure that the result is always positive, regardless of which side of the line the point lies on. If you get a negative value, it just means you forgot the absolute value bars. Always keep them in mind!

Question

Let the tangent to the parabola $y^2 = 12x$ at the point $(3, \alpha)$ be perpendicular to the line $2x + 2y = 3$. Then the square of distance of the point $(6, -4)$ from the normal to the hyperbola $\alpha^2 x^2 - 9y^2 = 9\alpha^2$ at its point $(\alpha - 1, \alpha + 2)$ is equal to .............

Sigma Percentile · Accepted Answer

### Analyzing the Parabolic Dance

We begin with the parabola $y^2 = 12x$. We are given a point $(3, \alpha)$ that lies on this curve.

The moment you see a point on a curve, your instinct should be to satisfy the equation. Substituting $x = 3$, we find:

$$\alpha^2 = 12(3) = 36 \Rightarrow \alpha = \pm 6$$

We need the tangent at this point to be perpendicular to the line $2x + 2y = 3$. Let us find the slope of the tangent by differentiating $y^2 = 12x$ with respect to $x$:

$$2y \frac{dy}{dx} = 12 \Rightarrow \frac{dy}{dx} = \frac{6}{y}$$

At our point $(3, \alpha)$, the slope $m_1$ is $\frac{6}{\alpha}$. The given line $2x + 2y = 3$ can be rewritten as $y = -x + \frac{3}{2}$, giving us a slope $m_2 = -1$.

Since the tangent is perpendicular to this line, we know $m_1 \cdot m_2 = -1$. Substituting our values:

$$\left( \frac{6}{\alpha} ight)(-1) = -1 \Rightarrow \alpha = 6$$

The ambiguity vanishes; our point is $(3, 6)$.

### The Hyperbolic Transformation

Now that we have unlocked $\alpha = 6$, the hyperbola equation $\alpha^2 x^2 - 9y^2 = 9\alpha^2$ becomes clear. Substituting $\alpha = 6$:

$$36x^2 - 9y^2 = 9(36) = 324$$

To see the true nature of this hyperbola, we divide by $324$, yielding:

$$\frac{x^2}{9} - \frac{y^2}{36} = 1$$

This is the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $a^2 = 9$ and $b^2 = 36$. We are given a point on this hyperbola: $(\alpha - 1, \alpha + 2)$, which becomes $(5, 8)$.

### The Geometry of the Normal

We need the normal to this hyperbola at $(5, 8)$. The standard formula for the normal to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$ is:

$$\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$$

Substituting $a^2 = 9$, $b^2 = 36$, $x_1 = 5$, and $y_1 = 8$:

$$\frac{9x}{5} + \frac{36y}{8} = 9 + 36$$

Simplifying this, we get $\frac{9x}{5} + \frac{9y}{2} = 45$. Dividing by $9$ and multiplying by $10$ gives us the equation:

$$2x + 5y - 50 = 0$$

### The Final Calculation

We need the distance from the point $(6, -4)$ to the line $2x + 5y - 50 = 0$. Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$:

$$d = \frac{|2(6) + 5(-4) - 50|}{\sqrt{2^2 + 5^2}} = \frac{|12 - 20 - 50|}{\sqrt{4 + 25}} = \frac{|-58|}{\sqrt{29}} = \frac{58}{\sqrt{29}}$$

Since $58 = 2 	imes 29$, this simplifies to $d = 2\sqrt{29}$. The question asks for the square of this distance, $d^2$:

$$d^2 = (2\sqrt{29})^2 = 4 	imes 29 = 116$$

The final answer is **116**.

Enter Numerical Value:

Visualized Solution

Analyze the Parabola $y^{2} = 12 x$

Slope of Tangent to Parabola

Perpendicularity Condition

Finding the Value of $α$

Equation of the Hyperbola

Point on the Hyperbola

Equation of the Normal

Simplifying the Normal Equation

Distance Formula Setup

Calculating the Distance

Final Answer: Square of Distance

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Parabolic Dance

The Hyperbolic Transformation

The Geometry of the Normal

The Final Calculation

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