Compilation of Problems in Analytic Geometry |

**COMPILATION OF MATH PROBLEMS**

**ANALYTIC GEOMETRY**

### PROBLEMS & SOLTUIONS:

**1. Find the equation of the directrix of the parabola y^2=16x.**

**Solution:**

The given equation is y^2 = 16x, which is the standard form of a parabola with vertex at the origin.

To find the equation of the directrix, we first need to determine the focus of the parabola. The focus is a point on the axis of symmetry that is equidistant from the vertex and the directrix.

The axis of symmetry is the vertical line passing through the vertex, which is x = 0.

The distance from the vertex (0,0) to the focus (p,0) is given by the equation p = 1/4a, where a is the coefficient of x in the equation of the parabola. In this case, a = 4, so p = 1.

Therefore, the focus is located at (1,0).

Now we can find the equation of the directrix. The directrix is a vertical line that is perpendicular to the axis of symmetry and is located at a distance of p units to the left of the vertex. Since the vertex is at the origin, the equation of the directrix is x = -p.

Substituting p = 1, we get the equation of the directrix as x = -1.

Therefore, the equation of the directrix of the parabola y^2 = 16x is x = -1

To find the equation of the directrix, we first need to determine the focus of the parabola. The focus is a point on the axis of symmetry that is equidistant from the vertex and the directrix.

The axis of symmetry is the vertical line passing through the vertex, which is x = 0.

The distance from the vertex (0,0) to the focus (p,0) is given by the equation p = 1/4a, where a is the coefficient of x in the equation of the parabola. In this case, a = 4, so p = 1.

Therefore, the focus is located at (1,0).

Now we can find the equation of the directrix. The directrix is a vertical line that is perpendicular to the axis of symmetry and is located at a distance of p units to the left of the vertex. Since the vertex is at the origin, the equation of the directrix is x = -p.

Substituting p = 1, we get the equation of the directrix as x = -1.

Therefore, the equation of the directrix of the parabola y^2 = 16x is x = -1

2. The midpoint of the line segment between P1 and P2 (-2, 4) is P (2, -1). Find P12. The midpoint of the line segment between P1 and P2 (-2, 4) is P (2, -1). Find P1

**Solution:**

Let P1 be (x1, y1) be the coordinates of the point P1.

The midpoint formula tells us that the midpoint of the line segment between two points (x1, y1) and (x2, y2) is given by:

((x1 + x2)/2, (y1 + y2)/2)

In this case, the midpoint of the line segment between P1 and P2 is (2, -1), so we can write:

((x1 + (-2))/2, (y1 + 4)/2) = (2, -1)

Simplifying this equation, we get:

(x1 - 2, y1 + 4) = (4, -2)

Equating the corresponding x and y coordinates, we get two equations:

x1 - 2 = 4

y1 + 4 = -2

Solving for x1 and y1, we get:

x1 = 6

y1 = -6

Therefore, P1 is (6, -6).

The midpoint formula tells us that the midpoint of the line segment between two points (x1, y1) and (x2, y2) is given by:

((x1 + x2)/2, (y1 + y2)/2)

In this case, the midpoint of the line segment between P1 and P2 is (2, -1), so we can write:

((x1 + (-2))/2, (y1 + 4)/2) = (2, -1)

Simplifying this equation, we get:

(x1 - 2, y1 + 4) = (4, -2)

Equating the corresponding x and y coordinates, we get two equations:

x1 - 2 = 4

y1 + 4 = -2

Solving for x1 and y1, we get:

x1 = 6

y1 = -6

Therefore, P1 is (6, -6).

**3. Given an ellipse: (x^2)/36 + (y^2)/32=1. Determine the distance between foci.**

**Solution:**

The equation of the ellipse is:

(x^2)/36 + (y^2)/32 = 1

We can write this equation in standard form by dividing both sides by the constant on the right side:

(x^2)/36/1 + (y^2)/32/1 = 1

Simplifying, we get:

(x^2)/36 = 1 - (y^2)/32

We can rewrite this equation as:

(x^2)/a^2 - (y^2)/b^2 = 1

where a^2 = 36 and b^2 = 32.

The distance between the foci of an ellipse is given by:

c = sqrt(a^2 - b^2)

Plugging in the values of a and b, we get:

c = sqrt(36 - 32) = sqrt(4) = 2

Therefore, the distance between the foci of the ellipse is 2.

(x^2)/36 + (y^2)/32 = 1

We can write this equation in standard form by dividing both sides by the constant on the right side:

(x^2)/36/1 + (y^2)/32/1 = 1

Simplifying, we get:

(x^2)/36 = 1 - (y^2)/32

We can rewrite this equation as:

(x^2)/a^2 - (y^2)/b^2 = 1

where a^2 = 36 and b^2 = 32.

The distance between the foci of an ellipse is given by:

c = sqrt(a^2 - b^2)

Plugging in the values of a and b, we get:

c = sqrt(36 - 32) = sqrt(4) = 2

Therefore, the distance between the foci of the ellipse is 2.

**4. Convert the θ = π/3 to the Cartesian equation.**

**Solution:**

The polar equation θ = π/3 describes a half-line starting at the origin and making an angle of π/3 with the positive x-axis in the counterclockwise direction.

To convert this to Cartesian coordinates, we can use the relationships between polar and Cartesian coordinates:

x = r cos(θ)

y = r sin(θ)

In this case, we have θ = π/3, so we can write:

x = r cos(π/3)

y = r sin(π/3)

Simplifying these equations, we get:

x = (1/2)r

y = (sqrt(3)/2)r

Now, we need to eliminate the parameter r to obtain the Cartesian equation. We can do this by squaring both sides of the equations for x and y, and then using the identity:

x^2 + y^2 = r^2

Substituting the expressions for x and y, we get:

(x^2 + y^2) = (1/4)r^2 + (3/4)r^2

Simplifying, we get:

x^2 + y^2 = r^2 = (4/3)(x^2 + y^2)

Multiplying both sides by 3/4, we get:

3x^2 + 3y^2 = 4x^2 + 4y^2

Simplifying further, we get:

x^2 - y^2 = 0

Therefore, the Cartesian equation of the polar equation θ = π/3 is x^2 - y^2 = 0.

To convert this to Cartesian coordinates, we can use the relationships between polar and Cartesian coordinates:

x = r cos(θ)

y = r sin(θ)

In this case, we have θ = π/3, so we can write:

x = r cos(π/3)

y = r sin(π/3)

Simplifying these equations, we get:

x = (1/2)r

y = (sqrt(3)/2)r

Now, we need to eliminate the parameter r to obtain the Cartesian equation. We can do this by squaring both sides of the equations for x and y, and then using the identity:

x^2 + y^2 = r^2

Substituting the expressions for x and y, we get:

(x^2 + y^2) = (1/4)r^2 + (3/4)r^2

Simplifying, we get:

x^2 + y^2 = r^2 = (4/3)(x^2 + y^2)

Multiplying both sides by 3/4, we get:

3x^2 + 3y^2 = 4x^2 + 4y^2

Simplifying further, we get:

x^2 - y^2 = 0

Therefore, the Cartesian equation of the polar equation θ = π/3 is x^2 - y^2 = 0.

**5. Find the coordinates of the point P(2, 4) with respect to the translated axis with origin at (1, 3).**

**Solution:**

To find the coordinates of the point P(2, 4) with respect to the translated axis with origin at (1, 3), we need to shift the origin of the coordinate system to (1, 3).

Let Q be the point obtained by shifting P to the new origin. Then, the coordinates of Q are given by:

Q(x', y') = P(x + 1, y + 3) = (2 + 1, 4 + 3) = (3, 7)

Therefore, the coordinates of the point P(2, 4) with respect to the translated axis with origin at (1, 3) are (3, 7).

Let Q be the point obtained by shifting P to the new origin. Then, the coordinates of Q are given by:

Q(x', y') = P(x + 1, y + 3) = (2 + 1, 4 + 3) = (3, 7)

Therefore, the coordinates of the point P(2, 4) with respect to the translated axis with origin at (1, 3) are (3, 7).

## Post a Comment