The given equation of the ellipse is:

$(x - 2)^2 + (y - 4)^2 = 1$

This is the equation of a circle centered at $(2, 4)$ with a radius of $1$, not an ellipse. The general equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center, and $r$ is the radius.

Now, you're asking about a gradient (slope) of 5. To interpret this, let's assume you want to find the equation of a line with slope 5 that is tangent to the circle.

1. The slope-intercept form of a line is:

$y = mx + c$

where $m = 5$ is the slope. So, the equation becomes:

$y = 5x + c$

To find the points of tangency, you would substitute $y = 5x + c$ into the equation of the circle and solve for $c$ such that the line touches the circle at exactly one point (indicating tangency).

Substituting into the circle equation:

$(x - 2)^2 + ((5x + c) - 4)^2 = 1$

This gives a quadratic equation in $x$. For the line to be tangent, the discriminant of this quadratic equation must be zero (which ensures exactly one solution, corresponding to tangency).

Would you like me to solve this equation fully and find the points of tangency?

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Here are 5 questions to extend understanding:

1. How do you find the general equation of an ellipse from a given problem?
2. What are the conditions for a line to be tangent to a circle or ellipse?
3. How does the slope of a line affect its equation and relationship to conic sections?
4. Can you describe the difference between an ellipse and a circle in terms of their equations?
5. What does the discriminant of a quadratic equation tell you about the number of solutions?

Tip: The discriminant of a quadratic equation, $ax^2 + bx + c = 0$, is $b^2 - 4ac$. When it's zero, the equation has exactly one solution (tangency in geometry).