If the line y = mx + 7√3 is normal to the hyperbola x²/24 - y²/18 = 1, then a value of m is (1) √5/2

Q: Wait, why do we reject the negative root here? Or do we?

Great observation! We actually don't reject it. The problem asks for 'a value' of $m$, and since the normal can exist at two symmetric points on the hyperbola, we get two possible values for $m$: $\pm \frac{2}{\sqrt{5}}$. Both are valid, but only one is listed in the options.

Q: Why can't we just use the condition of tangency $c^2 = a^2m^2 - b^2$?

That is a common trap! That condition is specifically for the tangent line $y = mx + c$. For a normal, the geometry is different. A normal is perpendicular to the tangent, so you must use the normal equation formula $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$ instead.

Q: How do we know the normal passes through $(x_1, y_1)$?

By definition, the normal to a curve at a point $P$ is the line perpendicular to the tangent at $P$. Therefore, it must pass through $P(x_1, y_1)$. This is why we can substitute $(x_1, y_1)$ into the hyperbola equation to solve for the coordinates.

Question

If the line $y = mx + 7\sqrt{3}$ is normal to the hyperbola $\frac{x^2}{24} - \frac{y^2}{18} = 1$, then a value of $m$ is

Options:

(A) $\sqrt{5}/2$

(B) $3/\sqrt{5}$

(C) $2/\sqrt{5}$

(D) $\sqrt{15}/2$

Sigma Percentile · Accepted Answer

### The Geometry of the Normal

Welcome, future engineer. Today, we are not just solving a problem; we are dancing with the geometry of the hyperbola.

Imagine you are standing on the curve of the hyperbola defined by:

$$\frac{x^2}{24} - \frac{y^2}{18} = 1$$

At any point $P(x_1, y_1)$ on this curve, there exists a unique line perpendicular to the tangent—the normal. Our goal is to find the slope $m$ of this normal line given its equation $y = mx + 7\sqrt{3}$.

This is a classic JEE Advanced challenge that tests your ability to bridge the gap between coordinate geometry and algebraic manipulation.

### Phase 1: Unlocking the Hyperbola

First, we must understand our terrain. The hyperbola is given in its standard form:

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

By comparing this to our equation $\frac{x^2}{24} - \frac{y^2}{18} = 1$, we immediately identify our parameters: $a^2 = 24$ and $b^2 = 18$.

These values are the DNA of our hyperbola; they dictate its shape, its foci, and, crucially, the behavior of its normals. Without these, we are lost, so keep them close.

### Phase 2: The Algebraic Bridge

Now, we invoke the standard equation of the normal at a point $P(x_1, y_1)$ on the hyperbola:

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

Substituting our known values, $a^2 + b^2 = 24 + 18 = 42$, the equation becomes:

$$\frac{24x}{x_1} + \frac{18y}{y_1} = 42$$

This is the theoretical normal. However, the problem gives us a specific line: $y = mx + 7\sqrt{3}$.

To compare these, we must align them. Rearranging the given line into the form $mx - y + 7\sqrt{3} = 0$ and our derived normal into:

$$\frac{24}{x_1}x + \frac{18}{y_1}y - 42 = 0$$

We see that since they represent the same line, their coefficients must be proportional. This is the moment of truth:

$$\frac{24/x_1}{m} = \frac{18/y_1}{-1} = \frac{-42}{7\sqrt{3}}$$

### Phase 3: The Final Calculation

Let's simplify the constant ratio:

$$\frac{-42}{7\sqrt{3}} = -2\sqrt{3}$$

Now, we set the $y$-coefficient ratio equal to this: $\frac{18/y_1}{-1} = -2\sqrt{3}$, which simplifies to $y_1 = 3\sqrt{3}$.

Since $P(x_1, y_1)$ lies on the hyperbola, we substitute $y_1$ back into the hyperbola equation:

$$\frac{x_1^2}{24} - \frac{(3\sqrt{3})^2}{18} = 1$$

This simplifies to $\frac{x_1^2}{24} - 1.5 = 1$, leading to $x_1^2 = 60$, or $x_1 = \pm 2\sqrt{15}$.

Finally, we return to the $x$-coefficient ratio: $\frac{24/x_1}{m} = -2\sqrt{3}$. Solving for $m$, we get:

$$m = -\frac{4\sqrt{3}}{x_1}$$

Substituting $x_1 = \pm 2\sqrt{15}$, we find the final result:

**$$m = \mp \frac{2}{\sqrt{5}}$$**

We have arrived at our destination. The elegance of this result—the way the coefficients cancel and the geometry aligns—is the true beauty of mathematics. Keep practicing, and soon, this intuition will be second nature.

Animated Solution for Mathematics - Conic Sections: If the line $y = m x + 73$ is normal to the hyperbola $\frac{x ^{2}}{24} - \frac{y ^{2}}{18} = 1$ , then a value of $m$ is

Select Answer:

Visualized Solution

Identify Hyperbola Parameters

General Equation of Normal

Substitute Known Values

Standardize Given Line

Compare Coefficients

Simplify the Constant Ratio

Solve for $y_{1}$

Substitute $y_{1}$ into Hyperbola Equation

Solve for $x_{1}^{2}$

Find $x_{1}$

Express $m$ in terms of $x_{1}$

Calculate Final Values of $m$

Conclude the Correct Option

The Sigma Insight: Equation of Tangent and Normal

The Geometry of the Normal

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line y=mx+73​ is normal to the hyperbola 24x2​−18y2​=1, then a value of m is