The foci of a hyperbola are (±2, 0) and its eccentricity is 3/2. A tangent, perpendicular to the lin

Q: Wait, why do we use the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and not $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$?

That is a fantastic question! The orientation of the hyperbola is dictated by the position of its foci. Since the problem states the foci are at $(\pm 2, 0)$, they lie on the x-axis. This tells us immediately that the transverse axis is horizontal, which corresponds to the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If the foci were on the y-axis, the equation would flip!

Q: How do I know which sign to choose in the tangent equation $y = mx \pm \sqrt{a^2m^2 - b^2}$?

This is the 'trap' of the problem! The $\pm$ sign represents the two possible tangents with the same slope, one touching the upper branch and one touching the lower branch. Because the problem restricts the point of contact to the first quadrant, we must analyze the coordinates of the point of contact $(x_1, y_1)$. By ensuring $x_1 > 0$ and $y_1 > 0$, we force a specific sign for the y-intercept $c$.

Q: Why can't we just use energy conservation or some other physics principle here?

While physics problems often involve energy conservation, this is a pure coordinate geometry problem. We are dealing with the rigid, elegant properties of conic sections. The 'energy' here is the algebraic consistency of the hyperbola's definition. Stick to the geometric properties of tangents and slopes; they are the most reliable tools in your kit for this specific challenge.

Question

The foci of a hyperbola are $(\pm 2, 0)$ and its eccentricity is $\frac{3}{2}$. A tangent, perpendicular to the line $2x + 3y = 6$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $x$- and $y$-axes are $a$ and $b$ respectively, then $|6a| + |5b|$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with the skeleton of our hyperbola. The problem grants us the foci at $(\pm 2, 0)$. In the language of coordinate geometry, the foci of a standard hyperbola

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

are located at $(\pm ae, 0)$. This is our first breakthrough; we immediately see that $ae = 2$.

We are also given the eccentricity $e = \frac{3}{2}$. Substituting $e$ into our equation, we find:

$$a \cdot \frac{3}{2} = 2 \Rightarrow a = \frac{4}{3}$$

Squaring this, we obtain $a^2 = \frac{16}{9}$.

To find $b^2$, we recall the fundamental identity $b^2 = a^2(e^2 - 1)$, which expands to $b^2 = (ae)^2 - a^2$. Since we know $ae = 2$ and $a^2 = \frac{16}{9}$, the calculation becomes:

$$b^2 = 4 - \frac{16}{9} = \frac{36 - 16}{9} = \frac{20}{9}$$

### The Tangent's Dance

Now, consider the line $2x + 3y = 6$. Rearranging it into the slope-intercept form $y = mx + c$, we get $3y = -2x + 6$, or $y = -\frac{2}{3}x + 2$. The slope of this line is $-\frac{2}{3}$.

The problem states our tangent is perpendicular to this line. The condition for perpendicularity is that the product of the slopes must be $-1$. If our tangent has slope $m$, then:

$$m \cdot \left(-\frac{2}{3}\right) = -1 \Rightarrow m = \frac{3}{2}$$

The general equation for a tangent to a hyperbola with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$. Plugging in our values:

$$y = \frac{3}{2}x \pm \sqrt{\frac{16}{9} \cdot \left(\frac{3}{2}\right)^2 - \frac{20}{9}}$$

Inside the square root, we have:

$$\frac{16}{9} \cdot \frac{9}{4} - \frac{20}{9} = 4 - \frac{20}{9} = \frac{16}{9}$$

The square root of $\frac{16}{9}$ is $\frac{4}{3}$. Thus, our potential tangents are $y = \frac{3}{2}x \pm \frac{4}{3}$.

### The First Quadrant Filter

The point of contact $(x_1, y_1)$ for a tangent $y = mx + c$ is given by:

$$\left(-\frac{a^2m}{c}, -\frac{b^2}{c}\right)$$

For the point to be in the first quadrant, both $x_1$ and $y_1$ must be positive. Since $a^2, b^2,$ and $m$ are positive, the only way for $x_1$ and $y_1$ to be positive is if $c$ is negative.

Therefore, we must choose the negative sign:

$$y = \frac{3}{2}x - \frac{4}{3}$$

### The Grand Finale

Rearranging $y = \frac{3}{2}x - \frac{4}{3}$ into the intercept form $\frac{x}{A} + \frac{y}{B} = 1$, we get $\frac{3}{2}x - y = \frac{4}{3}$. Dividing by $\frac{4}{3}$ yields:

$$\frac{x}{8/9} + \frac{y}{-4/3} = 1$$

Thus, the x-intercept is $a = \frac{8}{9}$ and the y-intercept is $b = -\frac{4}{3}$. The final step is to calculate $|6a| + |5b|$:

$$\left|6 \cdot \frac{8}{9}\right| + \left|5 \cdot \left(-\frac{4}{3}\right)\right| = \left|\frac{16}{3}\right| + \left|-\frac{20}{3}\right| = \frac{16}{3} + \frac{20}{3} = \frac{36}{3} = 12$$

The final result is **12**.

Enter Numerical Value:

Visualized Solution

The Hyperbola and its Foci

Finding the Semi-Major Axis $a$

Finding $b^{2}$

Calculating $b^{2}$

The Reference Line

Slope of the Tangent $m$

Equation of Tangent to Hyperbola

Substituting Values into Tangent Equation

Simplifying the Tangent Equation

Selecting the Correct Tangent

Finding the Intercepts $a$ and $b$

Final Calculation of $∣6 a ∣ + ∣5 b ∣$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Dance

The First Quadrant Filter

The Grand Finale

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