Parabola

The correct answer is (B) : 8
Vertex of Parabola : (2, –1)
and directrix : 4x – 3y = 21
Distance of vertex from the directrix

∴ length of latus rectum
= 4a
= 4 × 2= 8

The correct answer is (B) : y = 4(x – 1)
Equation of tangent of slope m to y = x²

Equation of tangent of slope m to y = –(x – 2)²

If both equation represent the same line

m = 0, 4
So, equation of tangent

The correct answer is 10
Focus = (4, 4) and vertex = (3, 2)
∴ Point of intersection of directrix with axis of parabola = A = (2, 0)
Image of A(2, 0) with respect to line
x + 2y = 6 is B(x₂, y2)


Point B is point of intersection of direction with axes of parabola P₂.
∴ x + 2y = λ must have point

∴ x + 2y = 10

The correct answer is (B) :
Let P(at², 2at) where
T :yt = x + at² So point Q is

N :y = –tx + 2at + at³ passes through (5, –8)

⇒ 3t³ – 4t + 16 = 0
⇒ (t + 2)(3t² – 6t + 8) = 0
⇒ t = –2
So ordinate of point Q is

The correct answer is (D) : any a > 0

Equation of tangent on parabola

y = 3x + 5

m-3 = 1+3m
m=-2
m-3 = -1-3m


Equation of one tangent :
Equation of other tangent :
Point of contact are

and
Now or = 0 [ S is the focus ]



Always true

The correct answer is 146.

We are given the equation .
We can rewrite this equation by completing the square for the terms:



This equation represents a parabola. We can compare it to the standard form of a parabola, , where:
,
,
.
The vertex of the parabola in the plane is . In the plane, the vertex is obtained by setting and , which gives and .
The focus of the parabola in the plane is , which is . In the plane, the focus is found by setting and , resulting in and . Therefore, the focus is .
Focal Chord
We are given a point on the parabola. A line passing through the focus can be written as:

where is the slope of the line.
Since the point lies on this line, we can substitute its coordinates into the equation:


So, the equation of the focal chord is:

Conclusion
The length of the focal chord is .

Parametric form and focal chord:
The parabola is . Its focus is . A chord through the focus is a focal chord. In the standard form with , points are . A known result: if corresponds to parameter and to then is a focal chord iff

Coordinates of and :
Let so that . Imposing and the condition that has positive ordinate, we arrive (from the given steps or direct calculation) at They do lie on , pass through , and are 100 units apart.

Point such that :
Since divides in the ratio , it is found using the section formula:

Line perpendicular to through :
The slope of is Hence a line perpendicular to has slope . The equation of the perpendicular through is: so

Checking given points: We substitute each point into :

: Matches.

: Matches.

: Matches.

: not

Hence does not lie on the required perpendicular, so it is the correct choice.

Given: The parabola is , and it passes through the points and .

• Step 1: Find the value of :

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• Thus, the equation of the parabola becomes:

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• Step 2: Area covered by the parabola:

The area under the parabola is given by:

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• Integrate:

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• Substitute the limits:

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• Step 3: Required area:

The required area is the area of the sector minus the area of the square minus the area under the parabola:

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• Step 4: Simplify the calculation:

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• Multiply through by 12 to simplify:

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Final Answer: The required area is .

We are given the parabola equation and the line equation . We need to find the length of the chord formed by the intersection of these two curves.
Step 1: Find the intersection of the parabola and the line.
Substitute the equation of the line into the equation of the parabola : Expanding the equation: Rearranging the terms: This is a quadratic equation in , and the points of intersection correspond to the roots of this equation.
Step 2: Use the centroid information.
The centroid is given as the centroid of the triangle . Using the centroid formula: where is the focus of the parabola and the points and are the points of intersection of the line and the parabola. This gives us two equations: 1. , 2. .
Step 3: Solve for and .
We are given that , so . To find the value of , use the relation derived from the intersection points. First, substitute into the equation of the parabola: Rearrange and simplify: Substitute into the equation: Now substitute into to find :
Step 4: Calculate the Points of Intersection.
Substitute and into the quadratic equation. The equation becomes: Solving this quadratic equation, we get the roots: For , substitute back into the line equation to get . Thus, point is one intersection point. For , substitute back into the line equation to get . Thus, point is the other intersection point.
Step 5: Calculate the Length of the Chord .
The distance between the points and is given by the distance formula:
Final Answer: 325.

Given the parabola:

The length of a focal chord is given by:

For this parabola, , so:

Solving for :

Thus:
Using the trigonometric identity :

The slope of the focal chord PQ is .

The equation of the chord passing through the focus is given by:

Simplifying:
To find the perpendicular distance of this line from the origin , use the formula:

Calculating :

To find the distance squared between the intersection points of the two parabolas, let's analyze each parabola:
1. Parabola with the x-axis as directrix: The standard form of a parabola with a focus at and directrix is , where is the distance from the focus to the directrix. Here, the focus is with the x-axis, or , as the directrix. This gives . Thus, the equation becomes .

2. Parabola with the y-axis as directrix: The standard form of a parabola with a focus at and directrix is . For this parabola, the focus is again and the directrix is . Therefore, , and the equation becomes .

To find the intersection points, solve the system of equations:

Consider solving for in terms of :
From , we get .

Substitute this into the second equation:

.

Simplify to get a polynomial equation:

Let . Then , leading to , which gives solutions or . So, or .

These yield and .

For , find by substituting back into :
Thus, point is .

Now, for :
.
The expression becomes

To find , compute .

Simplify the expressions using algebraic manipulation to reveal:

. So, .

The equation of a parabola is given by the definition: the distance from any point on the parabola to the focus is equal to the perpendicular distance from the point to the directrix.
Given:
- Focus
- Directrix:
The distance from the point on the parabola to the focus is: The distance from to the directrix is: For , we substitute into both expressions: Simplifying both sides, we solve for . After solving, we find:
Thus, the sum of the ordinates of the points on the parabola is .

The problem asks for the value of . We are given a parabola and its focal chord PQ which makes an angle of with the positive x-axis. A circle with diameter PS (S is the focus) touches the y-axis at . Point P lies in the first quadrant.

Concept Used:

The solution involves concepts from coordinate geometry, specifically parabolas and circles.

1. Parabola :
   • The focus is at the point .
   • Any point on the parabola can be represented in parametric form as .
2. Slope of a Line: The slope of a line segment joining points and is given by . If the line makes an angle with the positive x-axis, its slope is .
3. Circle with Diameter: For a circle whose diameter has endpoints and :
   • The center of the circle is the midpoint of the diameter: .
   • The radius is half the length of the diameter.
4. Circle Touching an Axis: A circle with center and radius touches the y-axis if . The point of tangency is .

Step-by-Step Solution:

Step 1: Find the focus of the parabola and the parametric coordinates of P.

The equation of the parabola is . Comparing this with the standard form , we get , which implies .

The focus of the parabola is at , so is at .

Let the coordinates of point P on the parabola be in parametric form: .

Step 2: Use the slope of the focal chord to find the parameter .

The focal chord PQ passes through P and S. The angle it makes with the positive x-axis is . Therefore, the slope of the chord PS is .

The slope of the line segment PS joining and is:

Equating the slope to :

Solving this quadratic equation for :

This gives two possible values for : and .

Since P lies in the first quadrant, its y-coordinate must be positive, which means . Thus, we choose .

The coordinates of P are .

Step 3: Find the center and radius of the circle with diameter PS.

The endpoints of the diameter are and .

The center of the circle, , is the midpoint of PS:

So, the center is .

The radius is the distance from the center to either P or S. Let's use S:

Step 4: Find the value of .

The circle with center and radius touches the y-axis. The condition for this is , which is satisfied as .

The point of tangency with the y-axis is given by .

In our case, the point of tangency is .

The problem states that this point is . Therefore, .

Final Computation & Result:

We are asked to find the value of .

The value of is 15.

Now, the parameter for point is . Therefore, the parameters for point are , which gives the coordinates of as . \textbf{Area of trapezium PQNM:}

The problem involves a geometric configuration with lines and curves, and we are tasked with finding the value of . Let's break down the solution in detail.

Step 1: Understanding the Given Information

The given equation is: This represents a line with slope and y-intercept . The points of intersection of the line with the axes are marked in the figure. - The line intersects the x-axis at . - The line intersects the y-axis at . The area in question is the area of a right triangle formed by the x-axis, y-axis, and the line .

Step 2: Area of the Triangle

The area of a triangle is given by the formula: In this case, the base of the triangle is the distance from the origin to , which is 4 units. The height is the distance from the origin to the point , which is also 4 units. Therefore, the area is: Hence, we can set .

Step 3: Solving for

The image shows the relationship between the variables and . Given the geometric configuration, we know that: Thus, the value of is 14.

Final Answer:

To find the angle that the line segment subtends at the vertex of the parabola, we first need to determine the points of intersection and where the line and the parabola intersect.

1. Rewrite the equation of the line and the parabola for ease of calculation:
   
   The equation of the parabola is .
2. Set the two expressions for equal to find the -coordinates of points and :
   
   Multiplying every term by 4 to eliminate fractions:
   
   Rearranging gives: .
3. Simplify and solve the quadratic equation:
   Divide the entire equation by 3: .
   Factorize: .
   The solutions are and .
4. Substitute back to find -coordinates:
   For , .
   For , .
5. Hence, the points A and B are and .
6. Determine the coordinates of the vertex of the parabola. For , the vertex is at the origin, .
7. Calculate the slopes of the lines and :
   • Slope of : .
   • Slope of : .
8. Calculate the angle between and using the formula for angle between two lines:
   Substituting the values, we get:
   Simplifying:

Therefore, the correct answer is .

The given parabola is , which is a standard parabola of the form where , so . The focus of the parabola is at .

Given is a focal chord and , for a parabola , if and , the points and will satisfy the conditions:

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Thus,

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Substituting :

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With , we have:

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This simplifies to , leading to:

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The focal chord condition further implies simple trigonometric identities and calculations for radii and midpoints, eventually leading us to the center and radius of the circle with equation:

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The circle's standard form must be derived implicitly from these parameters, giving:

and .

So, the expression results in:

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This calculated result, , perfectly fits the expected range of provided by the question.

The equation of the parabola is: The focus is , and the point is .
From the equation of the parabola, we know the parametric equations for the points on the parabola: Let divides internally in the ratio : Thus, the ratio , and:

Given that lies on the parabola , we can find by substituting the coordinates of P: Now, since PQ is a focal chord, we use the standard result for the area of a quadrilateral formed by the focal chord and perpendiculars from the points on the parabola to the directrix.

The area of quadrilateral PQMN is .