Instruction: Show all the necessary steps

Question

Instruction: Show all the necessary steps

1. Find the equation of the tangent line to the circle with equation (𝑥 − 1)
2 + (𝑦 − 1)
2 = 4
at point P(−1, −1).
2. Find the focus, vertex, directrix, length of latus rectum and axis of a parabola whose
equation is (𝑥 + 1)
2 = −𝑦 + 1.
3. Given centre C (8, 2); one focus F(4,2) and 0ne vertex (12, 2. Then find the equation of
the ellipse.
4. Given foci at F (0, 4) and F’ (0, -4); length of transverse axis is 4. Then find the equation
of the hyperbola and find its asymptotes.
5. Let 𝑎 = 2𝑖 + 3𝑗 𝑎𝑛𝑑 𝑏 = −3𝑖 − 𝑗 be vectors. Find a unit vector in the direction of the
vector 𝑎 − 2𝑏
6. If a translation T takes the (1, 1) to (–2, 2), then find
a. The images of a line −2𝑥 + 𝑦 − 7 = 0.
b. The image of the triangle with vertices A (0, –2), B (–1, 2) and C (-1, 1) .
c. The equation of the image for the circle whose equation is 𝑥
2 + 𝑦
2 − 4𝑦 − 1 = 11
7. Find the image of the point (–2, 4) after a reflection about the line y = x + 2.
8. Find the image of (2,-1) after it has been rotated −1200 about (1, 3)

Answer 1

1. To find the equation of the tangent line, first find the derivative of the circle's equation with respect to x:

2(x-1) + 2(y-1)dy/dx = 0
dy/dx = - (x-1)/(y-1)

Substitute the coordinates of point P into the derivative:
dy/dx = -(-1-1)/(-1-1) = 1

The slope of the tangent line is 1.
Since the tangent line passes through point P(-1, -1),
Equation of tangent line: y = x

2. For the parabola,
Vertex: (-1, 1)
Focus: (-1, 0)
Directrix: y = 2
Length of latus rectum: 1
Axis: x = -1

3. The equation of the ellipse is (x-8)^2/16 + (y-2)^2/36 = 1

4. For the hyperbola with foci at (0, 4) and (0, -4) and length of transverse axis 4,
Equation of hyperbola: y^2/4 - x^2/16 = 1
Asymptotes: y = 2x and y = -2x

5. To find a unit vector in the direction of a - 2b,
a - 2b = 2i + 3j - 2(-3i - j) = 8i + 5j
Unit vector: (8i + 5j) / sqrt((8)^2 + (5)^2)

6.
a. The image of the line -2x + y - 7 = 0 after translation will be -2x + y - 3 = 0
b. The image of the triangle after translation will be A' (1, 0), B' (0, 4) and C' (0, 3)
c. The equation of the image of the circle after translation will be x^2 + y^2 - 4y - 1 = 9

7. The image of the point (-2, 4) after reflection about the line y = x + 2 will be (4, -2)

8. The image of (2, -1) after rotation -120 degrees about (1, 3) will be (-5, 4).