Consider the hyperbola H : x² - y² = 1 and a circle S with center N(x₂, 0). Suppose that H and S tou

Q: Why does the normal to the hyperbola at $P$ necessarily pass through the center of the circle $N$?

Because they are tangent at $P$, the normal line is perpendicular to the common tangent. Since the normal to a circle always passes through its center, the normal to the hyperbola must also pass through $N$.

Q: I'm confused about the coordinates of $N$. How did we get $2x_1$?

Substitute $N(x_2, 0)$ into the normal equation $y - y_1 = -\frac{y_1}{x_1}(x - x_1)$. Setting $y=0$ and solving for $x_2$ yields $x_2 = 2x_1$. It's a beautiful simplification!

Q: Why can't we just differentiate $m$ with respect to $x_1$ directly without substituting $y_1$?

You can, but you must use the chain rule! Since $y_1 = \sqrt{x_1^2 - 1}$, you must account for the derivative of the square root function. Substituting first makes the algebra much cleaner.

Question

Consider the hyperbola $H : x^2 - y^2 = 1$ and a circle $S$ with center $N(x_2, 0)$. Suppose that $H$ and $S$ touch each other at a point $P(x_1, y_1)$ with $x_1 > 1$ and $y_1 > 0$. The common tangent to $H$ and $S$ at $P$ intersects the $x$-axis at point $M$. If $(l, m)$ is the centroid of the triangle $\Delta PMN$, then the correct expression(s) is(are)

Options:

(A) $\frac{dl}{dx_1} = 1 - \frac{1}{3x_1^2}$ for $x_1 > 1$

(B) $\frac{dm}{dx_1} = \frac{x_1}{3\sqrt{x_1^2 - 1}}$ for $x_1 > 1$

(C) $\frac{dl}{dx_1} = 1 + \frac{1}{3x_1^2}$ for $x_1 > 1$

(D) $\frac{dm}{dy_1} = \frac{1}{3}$ for $y_1 > 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The hyperbola is defined by the equation $H: x^2 - y^2 = 1$. We consider a point $P(x_1, y_1)$ on the right branch of this hyperbola, where $x_1 > 1$.

The tangent line to the hyperbola at point $P$ is given by the equation:
$$xx_1 - yy_1 = 1$$

This tangent line intersects the $x$-axis at point $M$. By setting $y=0$ in the tangent equation, we find $x = \frac{1}{x_1}$. Thus, the coordinates of $M$ are $(\frac{1}{x_1}, 0)$.

### The Tangent and the Normal

The slope of the tangent line at $P$ is $\frac{x_1}{y_1}$. Consequently, the slope of the normal line at $P$ is the negative reciprocal, which is $-\frac{y_1}{x_1}$.

Since the circle and the hyperbola are tangent at $P$, the normal to the hyperbola at $P$ must also be the normal to the circle. Because all normals to a circle pass through its center, we denote the center of the circle as $N(x_2, 0)$.

### Unveiling the Center $N$

The normal line passes through $P(x_1, y_1)$ and $N(x_2, 0)$. Using the point-slope form, the equation of the normal is:
$$y - y_1 = -\frac{y_1}{x_1}(x - x_1)$$

Substituting the coordinates of $N(x_2, 0)$ into this equation, we obtain:
$$0 - y_1 = -\frac{y_1}{x_1}(x_2 - x_1)$$

Since $y_1 > 0$, we divide by $-y_1$ to simplify the expression:
$$1 = \frac{x_2 - x_1}{x_1} \Rightarrow x_1 = x_2 - x_1 \Rightarrow x_2 = 2x_1$$

Thus, the center of the circle is located at $N(2x_1, 0)$.

### The Centroid Synthesis

We now identify the vertices of triangle $\Delta PMN$ as $P(x_1, y_1)$, $M(\frac{1}{x_1}, 0)$, and $N(2x_1, 0)$. The centroid $(l, m)$ is the average of these coordinates.

For the $x$-coordinate $l$:
$$l = \frac{x_1 + \frac{1}{x_1} + 2x_1}{3} = \frac{3x_1 + \frac{1}{x_1}}{3} = x_1 + \frac{1}{3x_1}$$

For the $y$-coordinate $m$:
$$m = \frac{y_1 + 0 + 0}{3} = \frac{y_1}{3}$$

### The Calculus of Motion

We differentiate $l$ with respect to $x_1$:
$$\frac{dl}{dx_1} = 1 - \frac{1}{3x_1^2}$$

Given $y_1^2 = x_1^2 - 1$, we have $y_1 = \sqrt{x_1^2 - 1}$. Therefore, $m = \frac{\sqrt{x_1^2 - 1}}{3}$.

Differentiating $m$ with respect to $y_1$ yields:
$$\frac{dm}{dy_1} = \frac{1}{3}$$

Finally, differentiating $m$ with respect to $x_1$ using the chain rule:
$$\frac{dm}{dx_1} = \frac{1}{3} \cdot \frac{1}{2\sqrt{x_1^2 - 1}} \cdot 2x_1 = \frac{x_1}{3\sqrt{x_1^2 - 1}}$$

Select Answer:

Visualized Solution

Visualizing the Hyperbola $H$

Equation of Tangent at $P$

Finding Point $M$ on the $x$ -axis

The Normal and Circle Center $N$

Equation of the Normal at $P$

Finding the Center $N$

Forming Triangle $Δ P M N$

Centroid $(l, m)$ of $Δ P M N$

Simplifying the Centroid Coordinates

Differentiating $l$ with respect to $x_{1}$

Differentiating $m$ with respect to $y_{1}$

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

Differentiating $m$ with respect to $x_{1}$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent and the Normal

Unveiling the Center $N$

The Centroid Synthesis

The Calculus of Motion

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Select Answer:

Visualized Solution

Visualizing the Hyperbola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

Equation of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Finding Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M on the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-axis

The Normal and Circle Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Equation of the Normal at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Finding the Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Forming Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔPMN

Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (l,m) of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔPMN