Problem 7
Question
The centroid of the triangle formed by the feet of the normals from the point \((h, k)\) to the parabola \(y^{2}+4 a x\) \(=0,(a>0)\) lies on (A) \(x\)-axis (B) \(y\)-axis (C) \(x=h\) (B) \(y=k\)
Step-by-Step Solution
Verified Answer
The centroid lies on the x-axis (Option A).
1Step 1: Understanding the Normal Equation
The given parabola is \(y^2 + 4ax = 0\). The equation of the normal from a point \((h, k)\) to the parabola \(y^2 = 4ax\) is given by \(y = mx - 2am - am^3\). The feet of the normal are points on this curve that satisfy these conditions.
2Step 2: Finding the Feet of the Normals
Solving the system of equations from the normal's conditions of being on the parabola and passing through \((h, k)\), we obtain three roots for \(m\): \(m_1, m_2, m_3\). The corresponding coordinates of the feet of the normals are \((am_i^2, -2am_i)\) for each \(m_i\).
3Step 3: Calculating the Centroid
The centroid \((G_x, G_y)\) of the triangle formed by the feet of the normals is given by averaging these coordinates: \[ G_x = \frac{am_1^2 + am_2^2 + am_3^2}{3} \] \[ G_y = \frac{-2am_1 - 2am_2 - 2am_3}{3} \]. Using the property of the roots, \(m_1 + m_2 + m_3 = 0\), simplify to \(G_y = 0\).
4Step 4: Verifying the Condition for the Centroid
Since \(G_y = 0\), the centroid lies on the \(x\)-axis. Thus, the correct answer is \(x\)-axis.
Key Concepts
Parabola NormalRoots of EquationCentroid CoordinatesGeometry of Conics
Parabola Normal
In geometry, a normal line to a curve is a line perpendicular to the tangent at a given point. For parabolas, normal lines have unique properties that aid in analyzing geometric shapes like triangles. The equation of the normal from a point \((h, k)\) to a standard parabola \(y^2 = 4ax\) is given by \(y = mx - 2am - am^3\). Here, \(m\) represents the slope of the normal. This equation helps calculate the feet of the normals, i.e., the points where these normals intersect the parabola.
- The role of the "normal" is crucial in finding points based on given conditions.
- Distinguishing normals helps us to reach coordinates needed for further calculations such as triangle formation.
Roots of Equation
In this exercise, we derive the roots of an equation associated with the normal line to the parabola. Solving the conditions from the foot of the normal leads to a cubic equation in \(m\). The roots \(m_1, m_2, m_3\) are essential to find the specific coordinates of the feet of the normals. Knowing these roots is crucial because they directly determine where the normals intersect with the parabola.
- The roots \(m_1, m_2, m_3\) show the variety of slopes determining the normals.
- They are used for calculating the x- and y-coordinates for the triangle vertices in our problem.
Centroid Coordinates
The centroid of a triangle is the point where its three medians intersect. It can be visualized as the center of mass if the triangle were made of a uniform material. For any triangle, this centroid has coordinates that are the average of the triangle’s vertices. For this problem, the vertices are the feet of the normals on the parabola.
- Calculate the centroid using: \[ G_x = \frac{am_1^2 + am_2^2 + am_3^2}{3} \]
- And for the y-coordinate: \[ G_y = \frac{-2am_1 - 2am_2 - 2am_3}{3} \]
Geometry of Conics
A conic section, or "conic," is a curve obtained as the intersection of a cone with a plane. The parabola is one of the simplest conics, characterized by its unique curvature and symmetry. Understanding the geometry of conics is fundamental to comprehend properties such as tangents and normals.
- The parabola is defined by its focus and directrix, resulting in its characteristic shape.
- Its axis of symmetry helps locate critical points and other geometric attributes.
Other exercises in this chapter
Problem 5
Coordinates of any point on the parabola, whose focus is \(\left(\frac{-3}{2},-3\right)\) and the directrix is \(2 x+5=0\) is given by (A) \(\left(2 t^{2}+2,2 t
View solution Problem 6
The mirror image of the directrix of the parabola \(y^{2}=\) \(4(x+1)\) in the line mirror \(x+2 y=3\) is (A) \(x=-2\) (B) \(4 y-3 x=16\) (C) \(3 x+4 y+16=0\) (
View solution Problem 8
If from a point, the two tangents drawn to the parabola \(y^{2}=4 a x\) are normals to the parabola \(x^{2}=4 b y\), then (A) \(a^{2}>8 b^{2}\) (B) \(b^{2}>8 a^
View solution Problem 9
If the focus of the parabola \((y-\beta)^{2}=4(x-\alpha)\) always lies between the lines \(x+y=1\) and \(x+y=3\), then (A) \(1
View solution