The focal chord to y² = 16x is tangent to (x-6)² + y² = 2, then the possible values of the slope of

Q: Wait, why is the focus $(4, 0)$?

The standard form of a parabola is $y^2 = 4ax$. Comparing $y^2 = 16x$ with $y^2 = 4ax$, we get $4a = 16$, so $a = 4$. The focus is $(a, 0)$, which is $(4, 0)$.

Q: Why do we use the distance formula?

A line is tangent to a circle if and only if the shortest distance from the center of the circle to the line is exactly equal to the radius. This is the most efficient way to relate the line's slope to the circle's geometry.

Q: What if the slope $m$ is undefined?

A vertical line passing through $(4, 0)$ would be $x = 4$. The distance from $(6, 0)$ to $x = 4$ is $|6 - 4| = 2$. Since the radius is $\sqrt{2}$, this line is not tangent. Thus, we only need to consider non-vertical lines with slope $m$.

Question

The focal chord to $y^2 = 16x$ is tangent to $(x-6)^2 + y^2 = 2$, then the possible values of the slope of this chord, are

Options:

(A) $\{-1, 1\}$

(B) $\{-2, 2\}$

(C) $\{-2, -1/2\}$

(D) $\{2, -1/2\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The parabola is defined by the equation $y^2 = 16x$. By comparing this to the standard form $y^2 = 4ax$, we identify $4a = 16$, which gives $a = 4$. Thus, the focus of the parabola is located at $(4, 0)$.

The circle is given by the equation $(x-6)^2 + y^2 = 2$. This circle has its center at $(6, 0)$ and a radius $r = \sqrt{2}$.

### The Line of Action

A focal chord is a line passing through the focus $(4, 0)$ with a slope $m$. Using the point-slope form $y - y_1 = m(x - x_1)$, we write the equation of the line as:

$$y - 0 = m(x - 4)$$

Rearranging this into the general form $Ax + By + C = 0$, we obtain:

$$mx - y - 4m = 0$$

### The Dance of Tangency

For the line to be tangent to the circle, the perpendicular distance $d$ from the center $(6, 0)$ to the line must equal the radius $r = \sqrt{2}$. The formula for the distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is:

$$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$

Substituting our specific values into this formula, we get:

$$\frac{|m(6) - (0) - 4m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$$

Simplifying the expression inside the absolute value, we have $|6m - 4m| = |2m|$. The equation becomes:

$$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$$

### The Algebraic Resolution

To solve for $m$, we square both sides of the equation to eliminate the absolute value and the square root:

$$\frac{4m^2}{m^2 + 1} = 2$$

Cross-multiplying yields:

$$4m^2 = 2(m^2 + 1)$$

$$4m^2 = 2m^2 + 2$$

Subtracting $2m^2$ from both sides results in $2m^2 = 2$, which simplifies to $m^2 = 1$. Taking the square root, we find the slopes:

**$m = \pm 1$**

These are the two possible slopes for the focal chords that are tangent to the given circle.

Animated Solution for Mathematics - Conic Sections: The focal chord to $y^{2} = 16 x$ is tangent to $(x - 6)^{2} + y^{2} = 2$ , then the possible values of the slope of this chord, are

Select Answer:

Visualized Solution

Identify the Parabola's Focus

Equation of the Focal Chord

Analyze the Circle

Apply Tangency Condition

Substitute Values into Distance Formula

Simplify the Numerator

Square Both Sides

Rearrange the Equation

Solve for Slope $m$

Conclusion and Summary

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Line of Action

The Dance of Tangency

The Algebraic Resolution

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The focal chord to y2=16x is tangent to (x−6)2+y2=2, then the possible values of the slope of this chord, are