Conic Sections

Step 1: Identify the center of the hyperbola
The foci of the hyperbola are given as and .
The center of the hyperbola is the midpoint of the foci:

Step 2: Find the values of and
The distance of each focus from the center is: The eccentricity is given as: Using the formula for eccentricity of a hyperbola: Substituting values: Solving for :

Step 3: Find the length of the transverse axis
The length of the transverse axis is given by:

Final Answer: The length of the transverse axis is 6.

Step 1: Identify the standard form of the ellipse
The standard form of the ellipse is: where is the center, is the semi-major axis, and is the semi-minor axis.
Given center: , so the equation becomes:

Step 2: Find the lengths of the semi-major and semi-minor axes
One vertex of the major axis is , so the distance from the center is: This is the semi-major axis, so and .
One vertex of the minor axis is , so the distance from the center is: This is the semi-minor axis, so and .

Step 3: Write the final equation
Substituting and into the equation:

Final Answer: The equation of the ellipse is .

The general equation of an ellipse is given by: Where is the center of the ellipse, is the semi-major axis, and is the semi-minor axis. The distance between the foci is given by , and we know that . From the given information, the endpoints of the major axis are and . The center of the ellipse is the midpoint of these two points, which is: Thus, the center of the ellipse is at . Now, the length of the major axis is the distance between and , which is . Therefore, the semi-major axis length is: Next, we know that the distance between the foci is 4, so: Now, we use the relationship to find . Finally, the equation of the ellipse is: Simplifying:

The correct option is (A) :

The equation of an ellipse with foci at and and semi-major axis and semi-minor axis is given by: where the relationship between the semi-major axis , semi-minor axis , and the distance of the foci is: We are given the foci at and , so the distance between the foci is: We are also given that the length of the major axis is 6, so: Now, using the equation , we can solve for : Thus, the length of the minor axis is:

The correct option is (D) :

We are given the equation of an ellipse: To find the centre of the ellipse, we need to rewrite this equation in the standard form of the ellipse equation: where is the centre of the ellipse.
Step 1: Group the -terms and -terms Rearrange the given equation: Now complete the square for both the -terms and -terms.
Step 2: Complete the square for the -terms Take the coefficient of , which is , divide it by 2, square it: Add 49 to both sides: which simplifies to:
Step 3: Complete the square for the -terms Take the coefficient of , which is , divide it by 2, square it: Add inside the bracket: Thus, the centre of the ellipse is .

The correct option is (C) :

The equation of the given parabola is .
This is a standard form of the parabola , where the focus is at .
Comparing with , we find , so .
Therefore, the focus of the parabola is at .
The line is given by the equation .
For this line to pass through the focus , substitute and into the line's equation:

The correct option is (B) :

The standard form of a hyperbola equation is where is the center, is the distance from the center to the vertices along the x-axis, and is the distance from the center to the vertices along the y-axis. The eccentricity is given by the formula: For the given hyperbola , we have and . Now, substitute these values into the eccentricity formula:

The correct option is (E) :

Since the major axis is along the x-axis, the equation of the ellipse is .

Substituting the given points:
For point (4, 3):
For point (-1, 4):

Adding the two equations:


From the second equation: . Substituting into the first equation:








Therefore .

Checking with the other equation: .

So the value of is 1.