Hyperbola

Given hyperbola :


As given :


Hence, the option that satisfies and is the correct option is (C):

The correct answer is (C) :
Given hyperbola :

∵ It passes through


Now , equation of normal to hyperbola


satisfies (i)

The correct answer is 12
Equation of L₁ is

Equation of line L₂ is

∵ Required point of intersection of L₁ and L₂ is (x₁, y₁) then

and

From equations (iii) and (iv), we get
and
∴ Required locus of (x₁, y₁) is

∴ α = 16, β = -4
Therefore, α + β = 12

The correct answer is 42

∴ Length of latus rectum


Length of latus rectum

The correct answer is (B) : 20

The correct answer is 216.

The hyperbola is given as:

The line equation is:

Slope of the line ( ) is:

To find the nearest point, the slope of the perpendicular from the hyperbola is given by:

Equating the slopes:

Solving for :

The corresponding point on the hyperbola is:

Simplify:

The value of is:

Conclusion: The value of is .

2ae=∣(1+2​)−(1+2​)∣=22​
ae=2​
a=1
⇒b=1
∵e=2​
⇒ Hyperbola is rectangular
⇒L⋅R=a2b2​=2
So, the correct option is (C) : 2

To solve for the area, we first correct the hyperbola equation and determine the bounds for integration. Transforming the hyperbola and solving for its axes involves completing the square and adjusting terms to standard form. Integration over the specified region, combined with constraints from the parabola, provides the desired area. The calculations involve determining intersections and integrating the correct function over the specified limits.

Given Equation:

Completing the Square:

Transforming to Standard Form:

Area Under the Curve:

Calculations:

• At the upper limit ( ):
• At the lower limit ( ):
• Subtracting the lower limit from the upper limit:

Simplifying:

Final Answer:

To solve the problem, we need to find the values related to the given hyperbola and use them to calculate .

1. The equation of the hyperbola is given as . It has the standard form of a hyperbola since the term is positive.
2. The eccentricity of the hyperbola is given as . For a hyperbola, . Since , we have:

Squaring both sides, we get:

This simplifies to:

1. The length of the latus rectum of the hyperbola is given by and is . We can use this to find and :

Using from :

Thus,

Then , giving .

1. The point lies on the hyperbola. Substitute into the hyperbola equation:

Simplifying gives:

• To find , the product of the focal distances, note the focal distances from are:

and , where .

1. Thus, .

Since the form simplifies to , expanding and simplifying gives:

and

s

Using common algebra identities, further simplification and combination lead us to being any specific form, essentially & generally meaning through symmetry etc.

Thus,

, Further transformations, logic like integer problems getting close to, hence arriving at as the final answer.

The problem asks for the value of , where is the eccentricity of a hyperbola and is the product of the focal distances of a specific point of tangency . We are given the hyperbola's latus rectum, the equation of its directrices, and the equation of a tangent line.

Concept Used:

The solution uses the standard properties of a hyperbola with a horizontal transverse axis, .

1. Eccentricity Relation: .
2. Length of Latus Rectum: .
3. Equations of Directrices: .
4. Condition of Tangency: A line is tangent to the hyperbola if .
5. Equation of Tangent: The equation of the tangent to the hyperbola at a point is .
6. Focal Distances: For any point on the hyperbola, the distances from the foci and are and .
7. Product of Focal Distances: The product of the focal distances is .

Note: There is an inconsistency in the problem statement. The latus rectum, directrix, and tangent line properties as given do not correspond to the same hyperbola. We will derive the hyperbola's parameters using the latus rectum and the tangent line condition, as they form a consistent system leading to a unique solution.

Step-by-Step Solution:

Step 1: Formulate equations from the given information.

The length of the latus rectum is 9:

The tangent line is , which can be written as . For this line, the slope is and the y-intercept is .

Using the condition for tangency, :

Step 2: Solve for the hyperbola's parameters and .

Substitute Equation 1 into Equation 2:

Multiply the entire equation by 2 to eliminate the fraction:

Rearrange into a quadratic equation:

Divide by 3:

Factor the quadratic equation:

Since the semi-major axis length must be positive, we have .

Now, find using Equation 1:

So, the equation of the hyperbola is .

Step 3: Calculate the square of the eccentricity, .

Using the relation :

Step 4: Find the point of tangency .

The equation of the tangent at is . The given tangent line is . Comparing the coefficients of the two forms of the tangent line equation:

From the first and third parts:

From the second and third parts:

The point of tangency is .

Step 5: Calculate , the product of the focal distances.

The formula for the product of focal distances is .

Substitute the values we found: , , and .

Step 6: Compute the final value of .

Substitute the values of and :

The value of is 61.