Conic Sections

To solve the problem, we need to evaluate the limit , where . Let's go through the steps:

1. Begin by analyzing the behavior of the numerator and the denominator as approaches 2.
2. We have in the numerator and in the denominator.
3. Substitute into :
   • , so becomes , which is not immediately obvious but impacts the indeterminate form.
4. In the denominator, as :
   • approaches .
5. The limit appears to be in an indeterminate form . Therefore, we apply L'Hôpital's Rule, which requires differentiating the numerator and the denominator:
6. Differentiate the numerator:
   • The derivative of is .
7. The derivative of the denominator is .

Apply L'Hôpital's Rule to evaluate the limit:

1. Substitute into the simplified expression:
   • and and multiply by zero, , to get the result: 1.

Thus, after simplifying and using L'Hôpital's Rule, we find:

The correct answer is .

To find the greatest value of the function on the interval , we need to perform the following steps:

1. Calculate the derivative of the function to find critical points:
   • The function is .
   • Using the product rule for differentiation, where and , we have:
   • .
   • Calculate .
   • Calculate .
   • The derivative is then: .
2. Set the derivative equal to zero to find critical points:
   • Solve .
   • Since for any real , we have , which gives .
3. Evaluate the function at the critical point and endpoints:
   • At : .
   • At : .
   • As , .
4. Conclusion:
   • The maximum value of on is , which occurs at .
   • Therefore, the correct answer is .

To determine the eccentricity of the conic given by the equation , we first need to identify the type of conic and express it in standard form.

1. Start with the given equation: .
2. Complete the square for the term:
3. Substitute back into the equation:
4. Divide the entire equation by 16 to convert it into standard form:
5. This equation represents an ellipse of the form :
   • Here, , , and .
   • For an ellipse, the eccentricity is given by .
6. Substitute and into the eccentricity formula:
7. Therefore, the eccentricity of the given conic is .

To solve the problem of finding the common tangents to the circle and the parabola , we need to understand the properties of tangents to conic sections and how they interact with each other.

The equation of the circle is , which is centered at the origin with a radius .

The equation of the parabola is , which opens to the right with its vertex at the origin.

The condition for common tangents to a circle and a parabola involves the geometric constructions that comply with both conics. We need to find a linear equation that simultaneously satisfies the tangent condition for the given circle and parabola.

First, consider the standard form of the tangent to the parabola . The equation of the tangent can be written as .

Next, for the circle , the standard form of the tangent is .

To have common tangents, the equations must be equivalent, requiring:

Squaring both sides and simplifying, this condition will yield the values of which satisfy both the equation of the circle's tangents and the parabola's tangents.

After solving the above conditions, the common tangent turns out to be of the form:

This means the line equations and are the common tangents to both the given circle and the parabola.

Thus, the correct answer is the option .