Let m1 and m2 be the slopes of the tangents drawn from the point P(4, 1) to the hyperbola H : y²/25

Q: Wait, why is it a vertical hyperbola? Shouldn't the $x^2$ term be positive?

That is a classic trap! In the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the $x^2$ term is indeed positive. However, look closely at our equation: $\frac{y^2}{25} - \frac{x^2}{16} = 1$. Here, the $y^2$ term is positive. This means the transverse axis is along the $y$-axis, making it a vertical hyperbola. Always check which variable is positive before assigning $a$ and $b$!

Q: Why do we square the tangent equation? Can't we just solve it as is?

The equation $1 = 4m \pm \sqrt{25 - 16m^2}$ contains a radical. To isolate $m$, we must eliminate the square root. Squaring both sides is the standard algebraic technique to remove the radical. Just be careful: when you square both sides, you might introduce extraneous roots, but in this specific geometry problem, the resulting quadratic correctly captures both possible tangent slopes.

Q: How do I know which sign to choose for the x-intercepts $\alpha$ and $\beta$?

The problem explicitly states that the tangents make *positive* intercepts on the $x$-axis. When you find the $x$-intercept by setting $y=0$ in the tangent equation $y = mx \pm c$, you will get two possible values for $x$. Simply choose the one that is positive to satisfy the condition $\alpha > 0$ and $\beta > 0$.

Question

Let $m_1$ and $m_2$ be the slopes of the tangents drawn from the point $P(4, 1)$ to the hyperbola $H : \frac{y^2}{25} - \frac{x^2}{16} = 1$. If $Q$ is the point from which the tangents drawn to $H$ have slopes $|m_1|$ and $|m_2|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$-axis, then $\frac{(PQ)^2}{\alpha\beta}$ is equal to _______.

Sigma Percentile · Accepted Answer

### The Geometry of Tangents: A Journey Through the Hyperbola

Welcome, fellow traveler, to the fascinating world of coordinate geometry. Today, we are going to dissect a problem that isn't just about numbers; it's about the elegant dance of lines and curves.

We are looking at the hyperbola $H : \frac{y^2}{25} - \frac{x^2}{16} = 1$. This is a vertical hyperbola, meaning its arms open upwards and downwards. The point $P(4, 1)$ sits outside this curve, watching it like a silent observer.

### Phase 1: The Tangent's Blueprint

To find the tangents, we need a tool. For any hyperbola of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$, the equation of a tangent with slope $m$ is given by:

$$y = mx \pm \sqrt{b^2 - a^2m^2}$$

Here, our $b^2 = 25$ and $a^2 = 16$. Substituting these, we get:

$$y = mx \pm \sqrt{25 - 16m^2}$$

This equation is our gateway. It tells us that for any slope $m$, there is a corresponding line that kisses the hyperbola perfectly.

### Phase 2: The Quadratic Trap

Since these tangents must pass through $P(4, 1)$, we substitute $x=4$ and $y=1$ into our tangent equation:

$$1 = 4m \pm \sqrt{25 - 16m^2}$$

To solve for $m$, we isolate the radical:

$$1 - 4m = \pm \sqrt{25 - 16m^2}$$

Squaring both sides gives us $(1 - 4m)^2 = 25 - 16m^2$. Expanding this, we get $1 + 16m^2 - 8m = 25 - 16m^2$.

Bringing everything to one side, we arrive at $32m^2 - 8m - 24 = 0$. Dividing by $8$, we find the beautiful, simple quadratic:

$$4m^2 - m - 3 = 0$$

Factoring this, we get $(4m + 3)(m - 1) = 0$. Thus, our slopes are $m_1 = 1$ and $m_2 = -\frac{3}{4}$.

### Phase 3: The Mystery of Point Q

Now, the problem takes a turn. We are introduced to a new point, $Q$, from which tangents are drawn with slopes $|m_1| = 1$ and $|m_2| = \frac{3}{4}$. We need to find the equations of these new tangents.

For $m=1$, the tangent is $y = x \pm 3$. Since the $x$-intercept must be positive, we set $y=0$ and find $x = \mp 3$. To get a positive intercept, we choose the line $y = x - 3$, where the intercept is $\alpha = 3$.

Similarly, for $m=\frac{3}{4}$, the tangent is $y = \frac{3}{4}x \pm 4$. Setting $y=0$, we get $x = \mp \frac{16}{3}$. For a positive intercept, we choose $y = \frac{3}{4}x - 4$, where the intercept is $\beta = \frac{16}{3}$.

### Phase 4: The Final Convergence

To find $Q$, we solve the system of our two new tangent lines: $y = x - 3$ and $y = \frac{3}{4}x - 4$. Equating them:

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

This simplifies to $\frac{1}{4}x = -1$, so $x = -4$. Substituting back, $y = -4 - 3 = -7$. Thus, $Q$ is at $(-4, -7)$.

Finally, we calculate $(PQ)^2$ using the distance formula between $P(4, 1)$ and $Q(-4, -7)$:

$$(PQ)^2 = (4 - (-4))^2 + (1 - (-7))^2 = 8^2 + 8^2 = 128$$

The product of the intercepts is $\alpha\beta = 3 	imes \frac{16}{3} = 16$. The ratio is:

$$\frac{(PQ)^2}{\alpha\beta} = \frac{128}{16} = 8$$

We have arrived at the finish line. The final result is **8**.

Enter Numerical Value:

Visualized Solution

Identify the Hyperbola $H$

General Equation of Tangent

Substitute Point $P (4, 1)$

Rearrange and Square

Form the Quadratic Equation

Solve for Slopes $m_{1}$ and $m_{2}$

Absolute Slopes for Point $Q$

Tangent with Slope $1$ and Intercept $α$

Tangent with Slope $\frac{3}{4}$ and Intercept $β$

Find Intersection Point $Q$

Calculate $(P Q)^{2}$

Final Ratio Calculation

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Tangents

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

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