Conic Sections

All equations resemble conic sections but checking the difference between them:

Let . Then Squaring and simplifying both sides leads to the equation of an ellipse. After algebraic simplification, it becomes:

To find the lengths of the axes of the conic, we rewrite the equation in standard form by completing the square.
After simplifying, we find the lengths of the axes to be 3 and 2.

The equation of the tangent to the ellipse can be written as . The distance from the center of the ellipse (which is at the origin) to this line is equal to the semi-major axis, which is 3. Using the formula for the distance from a point to a line: Thus: Since we are given that the line is tangent, the correct formula for is:

The general equation of a hyperbola is . The eccentricity of a hyperbola is given by: Comparing the given equation with the standard form, we find that the eccentricity is .

The equation of the parabola implies that the directrix is at . Solving for using the geometric properties of the parabola, we find .

The given equation is of the form: This is a second-degree equation in two variables. The presence of the term indicates that it is a parabola. Thus, the correct answer is Parabola.

We are given the curve and need to find a point where the tangent is parallel to the chord joining and . 1. Find the slope of the chord: The slope of the chord joining and is given by the formula: 2. Find the derivative of the curve: The derivative of the curve will give us the slope of the tangent at any point on the curve. Differentiate the given equation: 3. Set the derivative equal to the slope of the chord: Since the tangent is parallel to the chord, the slope of the tangent at the point of interest must be equal to the slope of the chord, which is 4: Solving for : 4. Find the corresponding -coordinates: Now, substitute into the original equation : Thus, the point on the curve is approximately , and the correct answer is (b) .