1. Question: Prove that the line passing through the points (2, 3) and (4, 5) is parallel to the line passing through the points (1, 2) and (3, 4).
Answer: To prove that the lines are parallel, we need to show that their slopes are equal. Let’s calculate the slopes of both lines. The slope of the first line is (5-3)/(4-2) = 1. The slope of the second line is (4-2)/(3-1) = 1. Since both slopes are equal, we can conclude that the lines are parallel.
2. Question: Find the equation of the circle passing through the points (1, 2), (3, 4), and (5, 6).
Answer: To find the equation of the circle passing through these points, we can use the general equation of a circle, which is (x-a)^2 + (y-b)^2 = r^2. Substitute the coordinates of the given points into this equation and solve the resulting system of equations to find the values of a, b, and r.
3. Question: Prove that the diagonals of a parallelogram bisect each other.
Answer: To prove that the diagonals of a parallelogram bisect each other, we can use the midpoint formula. Let the vertices of the parallelogram be A(x1, y1), B(x2, y2), C(x3, y3), and D(x4, y4). The midpoint of AC is ((x1+x3)/2, (y1+y3)/2), and the midpoint of BD is ((x2+x4)/2, (y2+y4)/2). If we can show that these two midpoints are the same, then it implies that the diagonals bisect each other.
4. Question: Find the equation of the line passing through the point (3, -2) and perpendicular to the line y = 2x + 1.
Answer: To find the equation of a line perpendicular to a given line, we need to determine its slope. The slope of the given line is 2. The slope of a line perpendicular to it will be the negative reciprocal of 2, which is -1/2. Now, we can use the point-slope form of a line to find the equation. The equation of the line passing through (3, -2) with a slope of -1/2 is y – (-2) = (-1/2)(x – 3).
5. Question: Prove that the distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is |c1 – c2| / sqrt(a^2 + b^2).
Answer: To prove this, we can use the formula for the distance between a point and a line. Let’s consider a point P(x0, y0) on the first line. The distance between the point P and the second line can be calculated using the formula |ax0 + by0 + c2| / sqrt(a^2 + b^2). Similarly, considering a point Q(x0, y0) on the second line, the distance between the point Q and the first line can be calculated using the formula |ax0 + by0 + c1| / sqrt(a^2 + b^2). By substituting the values of x0 and y0, we can show that the distances are equal to |c1 – c2| / sqrt(a^2 + b^2).
6. Question: Find the equation of the hyperbola with foci at (±2, 0) and vertices at (±5, 0).
Answer: The equation of a hyperbola centered at the origin with foci at (±c, 0) and vertices at (±a, 0) is given by x^2/a^2 – y^2/b^2 = 1. In this case, the distance between the foci is 2c = 4, and the distance between the vertices is 2a = 10. By solving these equations, we find that c = 2 and a = 5. Substituting these values into the equation, we get x^2/25 – y^2/4 = 1.
7. Question: Prove that the equation of the tangent to the ellipse x^2/a^2 + y^2/b^2 = 1 at the point (x1, y1) is xx1/a^2 + yy1/b^2 = 1.
Answer: To prove this, we can use the concept of the slope of a tangent to a curve. The slope of the tangent to the ellipse at the point (x1, y1) can be calculated as -b^2×1 / (a^2y1). Now, using the point-slope form of a line, we can write the equation of the tangent as y – y1 = (-b^2×1 / (a^2y1))(x – x1). Simplifying this equation, we get xx1/a^2 + yy1/b^2 = 1.
8. Question: Find the equation of the parabola with focus at (2, 3) and directrix y = -1.
Answer: The equation of a parabola with focus (a, b) and directrix y = c is given by (x – a)^2 = 4c(y – b). In this case, the focus is (2, 3) and the directrix is y = -1. By substituting these values into the equation, we get (x – 2)^2 = 4(-1)(y – 3), which simplifies to (x – 2)^2 = -4(y – 3).
9. Question: Prove that the equation of the normal to the curve y = f(x) at the point (x1, y1) is y – y1 = -1/f'(x1)(x – x1).
Answer: To prove this, we can use the concept of the derivative of a function. The slope of the tangent to the curve at the point (x1, y1) is given by f'(x1). The slope of the normal to the curve at the same point will be the negative reciprocal of f'(x1), which is -1/f'(x1). Now, using the point-slope form of a line, we can write the equation of the normal as y – y1 = -1/f'(x1)(x – x1).
10. Question: Find the equation of the ellipse with foci at (±3, 0) and major axis length 10.
Answer: The equation of an ellipse centered at the origin with foci at (±c, 0) and major axis length 2a is given by x^2/a^2 + y^2/b^2 = 1. In this case, the distance between the foci is 2c = 6, and the major axis length is 2a = 10. By solving these equations, we find that c = 3 and a = 5. Substituting these values into the equation, we get x^2/25 + y^2/16 = 1.