Let the eccentricity of the hyperbola x²/a² - y²/b² = 1 be 5/4. If the equation of the normal at the

Q: Wait, why do we use the normal formula $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$ instead of finding the slope of the tangent and then the normal?

You absolutely could find the slope of the tangent first, then take the negative reciprocal to find the slope of the normal, and finally use the point-slope form. However, in competitive exams like JEE, time is your most precious resource. The formula $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$ is a direct, derived result that saves you precious minutes and reduces the chance of algebraic errors. It is a powerful tool to keep in your arsenal.

Q: How do I know if the point $P$ is actually on the hyperbola? What if the question gives me a point that isn't on the curve?

That is a fantastic question! In a well-posed JEE problem, the point provided will always satisfy the equation of the curve. However, it is a great habit to verify this. If you ever suspect a point might not be on the curve, simply plug the coordinates into the hyperbola equation. If the left side equals the right side, you are safe to proceed. If it doesn't, double-check your calculations or the problem statement.

Q: Is there a difference between the normal to an ellipse and a hyperbola? I always get the signs mixed up.

It is very common to confuse these! Think of it this way: the hyperbola equation has a minus sign, $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The normal equation for the hyperbola has a plus sign, $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$. For the ellipse, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the normal equation has a minus sign, $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Just remember: the sign in the normal equation is the opposite of the sign in the curve's equation!

Question

Let the eccentricity of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\frac{5}{4}$. If the equation of the normal at the point $(\frac{8}{\sqrt{5}}, \frac{12}{5})$ on the hyperbola is $8\sqrt{5}x + \beta y = \lambda$, then $\lambda - \beta$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Geometry of the Hyperbola: A Journey into the Normal

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are exploring the elegant architecture of the hyperbola.

Imagine standing on the coordinate plane, looking at the two graceful, sweeping branches of a hyperbola. We are given a specific point $P(\frac{8}{\sqrt{5}}, \frac{12}{5})$ resting on one of these branches. Our mission is to find the equation of the normal line at this point and extract the values of $\beta$ and $\lambda$.

### Phase 1: Unlocking the Hyperbola's Dimensions

Every hyperbola is defined by its eccentricity, $e$. We are told $e = \frac{5}{4}$. This number is not just a ratio; it is the DNA of the hyperbola.

We know the fundamental relationship for a hyperbola is $b^2 = a^2(e^2 - 1)$. Substituting $e = \frac{5}{4}$, we get $e^2 = \frac{25}{16}$.

Thus, we find:
$$b^2 = a^2 \left( \frac{25}{16} - 1 ight) = a^2 \left( \frac{9}{16} ight)$$

This gives us our first crucial link: $b^2 = \frac{9}{16}a^2$. We have effectively reduced our two unknowns, $a^2$ and $b^2$, to a single variable, $a^2$.

### Phase 2: The Point of Contact

Now, we turn our attention to the point $P(\frac{8}{\sqrt{5}}, \frac{12}{5})$. This point is a prisoner of the hyperbola's equation.

By substituting $x = \frac{8}{\sqrt{5}}$ and $y = \frac{12}{5}$ into the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we obtain:
$$\frac{64}{5a^2} - \frac{144}{25b^2} = 1$$

Now, substitute our earlier finding, $b^2 = \frac{9}{16}a^2$, into this equation. The term $\frac{144}{25b^2}$ becomes:
$$\frac{144}{25 \left( \frac{9}{16}a^2 ight)} = \frac{144 	imes 16}{25 	imes 9a^2} = \frac{256}{25a^2}$$

Our equation now simplifies to:
$$\frac{64}{5a^2} - \frac{256}{25a^2} = 1 \Rightarrow \frac{320 - 256}{25a^2} = 1 \Rightarrow \frac{64}{25a^2} = 1$$

Thus, $a^2 = \frac{64}{25}$. Consequently, $b^2 = \frac{9}{16} 	imes \frac{64}{25} = \frac{36}{25}$. We have successfully decoded the dimensions of our hyperbola.

### Phase 3: The Normal Line

With $a^2 = \frac{64}{25}$ and $b^2 = \frac{36}{25}$, we are ready to construct the normal. The normal line at $(x_1, y_1)$ is given by:
$$\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$$

Substituting our values, we have:
$$\frac{\frac{64}{25}x}{\frac{8}{\sqrt{5}}} + \frac{\frac{36}{25}y}{\frac{12}{5}} = \frac{64}{25} + \frac{36}{25}$$

Simplifying the coefficients, the first term becomes $\frac{8\sqrt{5}}{5}x$ and the second term becomes $\frac{3}{5}y$. The right side is $\frac{100}{25} = 4$.

Multiplying the entire equation by $5$ to clear the denominators, we get:
$$8\sqrt{5}x + 3y = 20$$

Wait, let us re-evaluate the coefficient simplification: $\frac{64/25}{8/\sqrt{5}} = \frac{64}{25} \cdot \frac{\sqrt{5}}{8} = \frac{8\sqrt{5}}{25}$. Multiplying by $25$ yields $8\sqrt{5}x + 15y = 100$.

Comparing this to $8\sqrt{5}x + \beta y = \lambda$, we find $\beta = 15$ and $\lambda = 100$. The final answer, $\lambda - \beta$, is **$100 - 15 = 85$**. You have conquered the hyperbola!

Enter Numerical Value:

Visualized Solution

Visualizing the Hyperbola and Point $P$

Relating $a^{2}$ and $b^{2}$ via Eccentricity

Using the Point $P$ on the Hyperbola

Substituting $b^{2}$ in terms of $a^{2}$

Simplifying the Equation

Solving for $a^{2}$

Calculating $b^{2}$

Equation of the Normal

Substituting Values into Normal Equation

Simplifying the Normal Equation

Standardizing the Normal Equation

Comparing and Finding $λ - β$

The Sigma Insight: Equation of Tangent and Normal

The Geometry of the Hyperbola

Phase 1

Phase 2

Phase 3

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