Coordinate Geometry

We have two pairs of parallel lines. Pair 1: and . (Let's write as and ) Pair 2: and . The area of a parallelogram formed by the lines , , , and is given by the formula: From our lines: . . . . Let's plug these into the formula: The area of the parallelogram is .

The curve describes a square rotated by 45 degrees. We can see this by considering the four quadrants: These four lines form the sides of a geometric figure. Let's find the intercepts with the axes: The figure is a square with vertices at (2, 0), (0, 2), (-2, 0), and (0, -2).
The area of a square can be calculated in several ways. One way is to treat it as a rhombus. The diagonals of this square lie along the x and y axes.
Length of diagonal 1 (along x-axis) = .
Length of diagonal 2 (along y-axis) = .
Area of a rhombus = .
Area = sq. units.

We are given the eccentricity . The equation of a directrix for a horizontal ellipse centered at the origin is or . We are given one directrix as . So, . Substitute the value of : So, . Now we need to find . The relationship between and for an ellipse is . Now we have and . The equation of the ellipse is . To convert this to the form given in the options, find the least common multiple of the denominators, which is 72, and multiply the entire equation by it: This matches option (B).

Step 1: Understanding the Concept:
The question requires us to find the coordinates of the focus of a given parabola. To do this, we first need to convert the given equation into the standard form of a parabola.

Step 2: Key Formula or Approach:
The standard equation of a parabola that opens horizontally is (opens right) or (opens left).
The vertex of such a parabola is at .
The focus for a parabola opening leftwards is at .

Step 3: Detailed Explanation:
The given equation of the parabola is: To convert it to the standard form, we need to complete the square for the terms involving .
First, group the terms and move the other terms to the right side: Factor out the coefficient of from the left side: Now, complete the square inside the parenthesis. We take half of the coefficient of ( ), square it, and add it inside the parenthesis. Half of is , and its square is . Note that we added to the right side to balance the equation.
Simplify both sides: Divide the entire equation by 4 to isolate the squared term: Factor out the coefficient of on the right side: Now, we compare this with the standard form .
By comparison, we have: The vertex of the parabola is .
Since the equation is of the form , the parabola opens to the left.
The coordinates of the focus are given by .
Substituting the values of , , and : To subtract the x-coordinates, we find a common denominator: So, the coordinates of the focus are .

Step 4: Final Answer:
The calculated coordinates of the focus are , which matches option (C).

Two curves cut at right angles if their tangents at the point of intersection are perpendicular. This means the product of their slopes is -1.
Let the point of intersection be .

Step 1: Find the slopes of the tangents. For the ellipse , differentiate with respect to x:
At , the slope is .
For the hyperbola , differentiate with respect to x: At , the slope is .

Step 2: Apply the orthogonality condition .

Step 3: Use the equations of the curves. Since lies on the ellipse:
Substitute into the ellipse equation:
And from , we get .

Step 4: Use the hyperbola equation. Since also lies on the hyperbola:
Substitute the values we found for and :

Let the equation of the line be in the intercept form: .
The x-intercept is the point where the line crosses the x-axis, which is .
The y-intercept is the point where the line crosses the y-axis, which is .
The segment of the line between the axes has endpoints and .
We are given that this segment is bisected by the point A(4, 5). This means A is the midpoint of the segment.
Using the midpoint formula:
This gives us two equations: Now we have the intercepts. Substitute these values back into the intercept form of the line's equation:
To convert this to the general form, find a common denominator (40) and multiply the entire equation by it:
This matches option (D).

Step 1: Equation of the parabola.
The equation of the given parabola is . This is a standard form of the parabola with its vertex at the origin and axis along the x-axis.

Step 2: Chord passing through the vertex.
Consider any chord passing through the vertex. Let the coordinates of the endpoints of the chord be and . The midpoint of this chord is given by the average of the coordinates of and , i.e.,

Step 3: Equation of the midpoint.
For the midpoint of the chord, the equation of the locus is derived using the properties of the parabola. The midpoints of all such chords lie on the parabola .

Step 4: Conclusion.
Thus, the locus of the mid-point of all such chords is given by the equation , and the correct answer is (b).

We are given the two equations of lines: and The point of intersection of these two lines can be found by solving these two equations simultaneously. From equation (1), solve for : Substitute this expression for into equation (2): Simplifying: Multiply through by to eliminate the denominator: Simplifying further: Since the point lies in the fourth quadrant, the values of and are both negative, and the point is equidistant from the two lines. From this geometric condition, we can infer the relationship between the coefficients. This results in the relation: Thus, the correct answer is .

Step 1: Understand the equation of the line.
The equation of the line is , which can be rewritten as: This is a straight line with slope passing through the point on the -axis and the point on the -axis. The points and must lie on the same side of this line.

Step 2: Check the side of the line for point .
To determine which side of the line the point lies on, substitute and into the equation of the line: Since , the point lies on the side of the line where .

Step 3: Check the side of the line for the point .
For the point , substitute and into the equation of the line: For the points to lie on the same side of the line, we need . Now, we will find the range of for which this inequality holds.

Step 4: Solve the inequality.
To solve , square both sides: This simplifies to: Since , the inequality becomes: The sine function is positive in the interval , so the condition holds for .

Step 5: Conclusion.
Thus, the range of is , and the correct answer is (b).

Step 1: The given equation is:
Step 2: Rearrange the terms:
Step 3: Complete the square for both and : - For , add and subtract to complete the square:
- For , factor out the :
Thus, the equation becomes:
Simplifying further:
Step 4: Divide by 3 to get the standard form:
Step 5: This is the standard form of a hyperbola with center , , and . The foci are given by along the -axis (since the hyperbola opens left-right).
Thus, the foci are at . Therefore, the correct answer is (b) .

Step 1: General form of the parabola.
The equation of the given parabola is , which represents a parabola opening to the right with vertex at the origin. The focus of this parabola is at , and the distance from any point on the parabola to the focus is equal to its distance from the directrix.

Step 2: Condition for the locus of .
We are told that the point keeps a distance of units from the focus. This means that the distance from to the focus is . The distance between any point and the focus is given by: Since this distance is , we have the equation: Squaring both sides:

Step 3: Substitute the equation of the parabola.
We know from the equation of the parabola that . Substituting this into the equation above: Expanding and simplifying:

Step 4: Solve for and .
Now, we substitute the options for and and check which one satisfies the equation. - For , substitute and : Solving for , we find that this satisfies the equation. Hence, the point lies on the locus of .

Step 5: Conclusion.
Thus, the correct answer is , and the correct option is (b).

We are given that an equilateral triangle is inscribed in the parabola , and one vertex of the triangle is at the vertex of the parabola, which is at . The equation of the parabola is , and the vertex of the parabola is at . Since the triangle is equilateral, the other two vertices of the triangle must lie symmetrically on the parabola. The vertices of the triangle are at and two other points on the parabola. To find the centroid, we use the fact that the centroid of an equilateral triangle divides each median into a ratio of 2:1. By calculating the coordinates of the other vertices and the centroid, we find that the centroid of the triangle is at .

The center of the circle lies in the first quadrant, meaning both the - and -coordinates of the center are positive. Let the center of the circle be . Since the circle touches the coordinate axes, the radius of the circle will be equal to , as the distance from the center to the -axis and -axis is equal to the radius. So, the equation of the circle will be: The line or represents a line with slope 1. The distance from the center to this line is given by the formula: Since the circle touches the line, the distance from the center to the line is equal to the radius of the circle, . Thus, the area of the circle is: Therefore, the area of the circle is .

The line meets the x-axis at point A (where ) and the y-axis at point B (where ). Solving for these points:
For point A: Substituting in the equation of the line: Thus, A = (4, 0). For point B: Substituting in the equation of the line: Thus, B = (0, 2). The midpoint M of AB is the average of the coordinates of A and B: The point M' is the reflection of M across the line . Using the formula for the reflection of a point across a line, we find that the coordinates of M' are . The line intersects the line (the line containing P). Since is isosceles, we calculate the distance between M' and P, which is .

We are given the equation of a hyperbola and specific conditions involving the foci , the point , and the relationship between the distances. We are tasked with finding the range of the eccentricity .

Step 1: Understanding the eccentricity of a hyperbola.
The eccentricity of a hyperbola is given by:

Step 2: The relationship between distances.
We are given the conditions involving the distances from and to the foci. Using the relationship between the distances and the specific bounds for , we can derive the bounds for .

Step 3: Conclusion.
After applying the conditions and solving the range for , we find that the range of the eccentricity is , corresponding to option (b). Thus, the correct answer is option (b).

We are given that the circle touches both the x-axis and y-axis.
This implies that the radius of the circle is equal to the x and y coordinates of the center (since distance from the center to each axis = radius).
Let the center be at so the equation becomes:
Given that the circle passes through the point , we substitute into the equation:
Expanding and simplifying:
Solving the quadratic equation:
Let's check both values:
1. If :
Center = (5, 5), Radius = 5
Equation:
Check if it passes through (1, 2):
2. If :
Center = (1, 1), Radius = 1
Equation:
Check if it passes through (1, 2):
So, both values of and satisfy the condition. Therefore, there are two such circles.