General Equation Of A Line

Step 1: Identify the normal vector of the plane and the direction vector of the line .
Step 2: Calculate the angle between the line and the plane using the dot product: Substitute values: Step 3: Since the problem is about the angle between the line and the plane, we calculate the complementary angle . Therefore, the sine function is used:

Step 1: Start with the equation of the parabola .
Step 2: Use the point-slope form of the equation of a tangent to a parabola. The general form of the tangent line to a parabola is given by: For the given parabola , we compare it with and get , which gives . Substituting into the tangent formula, we get: Step 3: Substitute the point into the equation of the tangent line to find . We substitute and into the equation : Multiplying through by to clear the denominator: Rearranging the equation: Factoring the quadratic equation: Thus, or .
Step 4: Since satisfies the equation, substitute back into the tangent equation. We get:

We are given the cubic equation: and the roots of this equation are in arithmetic progression. Let the roots of the equation be , , and , where is the middle root and is the common difference.
Step 1: By Vieta's formulas, we know the following relationships between the roots and the coefficients of the cubic equation : - The sum of the roots is , - The sum of the products of the roots taken two at a time is ,
- The product of the roots is . For the equation , we have , , , and .
Step 2: From Vieta's formulas: 1. The sum of the roots is: Thus, . 2. The sum of the products of the roots taken two at a time is: Using Vieta’s formula: Substitute : Solving for :
Step 3: The product of the roots is: Using Vieta’s formula: Substitute and : This simplifies to: Multiplying both sides by 4:
Conclusion: Thus, the value of is .

We are given the quadratic equation: This is a quadratic equation of the form , where: The roots of a quadratic equation are given by the quadratic formula:
Step 1: First, calculate the discriminant : Substitute the values of , , and : Simplifying the discriminant: This discriminant is non-negative, indicating real roots.
Step 2: Now, using the quadratic formula, we can find the roots of the equation: After simplifying, the roots of the quadratic equation are:
Conclusion: Thus, the roots of the equation are .

To find the equation of the plane containing the lines and which pass through the origin, we use the cross product of their direction vectors.
Step 1: The direction vectors of and are and , respectively.
Step 2: Compute the cross product to find a normal to the plane: Step 3: The equation of the plane will be , since the normal vector to the plane is . Now, simplify by dividing the entire equation by 8 to reduce the coefficients: Thus, the equation of the plane is .

We are given the equation: Let the roots of this equation be . The equation can then be written as: We are asked to find the equation whose roots are the translates of these roots by . This means the new roots will be .
Step 1: To translate the roots by , we substitute for in the original equation. This gives the new equation: We now expand this expression by substituting into the equation .
Step 2: Substitute into the original equation: Now expand each term: Thus, the expanded equation is: Now combine like terms: This simplifies to: Thus, the required equation is .