If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equatio

Q: Wait, why do we use $e^2 = 1 + \frac{b^2}{a^2}$ instead of $e^2 = 1 - \frac{b^2}{a^2}$?

That is a classic confusion! Remember, for an ellipse, the eccentricity is less than one, so the formula involves subtraction: $e^2 = 1 - \frac{b^2}{a^2}$. But for a hyperbola, the curve is 'open' and stretches infinitely, meaning $e > 1$. Therefore, we must add the ratio of the squares of the axes to keep the eccentricity greater than one. It is all about the geometry of the curve!

Q: Why can't we just use calculus to find the slope of the tangent?

You absolutely can! You could differentiate the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ implicitly to find $\frac{dy}{dx}$ at $(4, 6)$ and then use the point-slope form. However, in the JEE, time is your most precious resource. The $T=0$ method is a specialized, elegant shortcut designed specifically for conic sections. It saves you from the tedious steps of differentiation and is much less prone to algebraic errors.

Q: Is the $T=0$ method applicable to all conic sections?

Yes, it is! Whether it is a circle, parabola, ellipse, or hyperbola, the $T=0$ rule is a universal tool. For any second-degree curve $f(x, y) = 0$, the tangent at $(x_1, y_1)$ is found by replacing $x^2$ with $x x_1$, $y^2$ with $y y_1$, $x$ with $\frac{x+x_1}{2}$, and $y$ with $\frac{y+y_1}{2}$. It is one of the most powerful 'hacks' in coordinate geometry.

Question

If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is-

Options:

(A) $2x - y - 2 = 0$

(B) $3x - 2y = 0$

(C) $2x - 3y + 10 = 0$

(D) $x - 2y + 8 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The hyperbola is a curve that stretches infinitely, defying the boundaries of a closed loop. To master the art of finding the tangent at the point $(4, 6)$, we must first decode the underlying architecture of the curve.

### Decoding the Eccentricity

Our journey begins with the eccentricity, $e=2$. In the world of conics, eccentricity is the DNA of the curve. For a hyperbola, the relationship between the semi-major axis $a$ and the semi-minor axis $b$ is governed by the elegant equation:

$$e^2 = 1 + \frac{b^2}{a^2}$$

Given $e=2$, we substitute this into our equation:

$$2^2 = 1 + \frac{b^2}{a^2} \Rightarrow 4 = 1 + \frac{b^2}{a^2}$$

With a simple rearrangement, we find that $\frac{b^2}{a^2} = 3$, or more simply, $b^2 = 3a^2$. This relationship is the bridge that connects the two axes, allowing us to define the entire hyperbola using only one parameter, $a^2$.

### The Geometry of the Point

Now that we have the relationship $b^2 = 3a^2$, we return to the standard equation of the hyperbola:

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

By substituting our expression for $b^2$, we obtain:

$$\frac{x^2}{a^2} - \frac{y^2}{3a^2} = 1$$

We are told the hyperbola passes through the point $(4, 6)$. This means the coordinates $x=4$ and $y=6$ must satisfy the equation:

$$\frac{4^2}{a^2} - \frac{6^2}{3a^2} = 1$$

Calculating the squares, we have $\frac{16}{a^2} - \frac{36}{3a^2} = 1$. Simplifying the second term, $36$ divided by $3$ is $12$, yielding:

$$\frac{16}{a^2} - \frac{12}{a^2} = 1 \Rightarrow \frac{4}{a^2} = 1$$

This tells us that $a^2 = 4$. Consequently, finding $b^2$ is trivial: $b^2 = 3(4) = 12$. Our hyperbola is now fully revealed as:

$$\frac{x^2}{4} - \frac{y^2}{12} = 1$$

### The Elegant Tangent

Finally, we arrive at the climax of our problem: finding the tangent at $(4, 6)$. We will use the JEE-favorite **$T=0$ method**, a powerful shortcut for finding tangents.

For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the tangent at $(x_1, y_1)$ is given by:

$$\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$$

Substituting our values, we get:

$$\frac{x(4)}{4} - \frac{y(6)}{12} = 1$$

The fours cancel out, leaving $x$, and $6$ over $12$ simplifies to $\frac{1}{2}$, leaving $\frac{y}{2}$. Thus, we have:

$$x - \frac{y}{2} = 1$$

Multiplying the entire equation by $2$ to clear the fraction, we arrive at the final result:

**$2x - y - 2 = 0$**

Animated Solution for Mathematics - Conic Sections: If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is-

Select Answer:

Visualized Solution

Standard Hyperbola Setup

Eccentricity Formula

Substituting Eccentricity

Relation Between $a^{2}$ and $b^{2}$

Single Variable Hyperbola

Substituting Point $P (4, 6)$

Solving for $a^{2}$

Finding $b^{2}$ and the Equation

Tangent Formula ( $T = 0$ )

Substituting into $T = 0$

Simplifying the Tangent Equation

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Decoding the Eccentricity

The Geometry of the Point

The Elegant Tangent

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