Problem 8
Question
A straight line is always parallel to the \(y z\) plane and intersects the curves \(x^{2}+y^{2}=a^{2}, z=0\) and \(x^{2}=a z, y=0\). Prove that it generates the surface \(x^{4} y^{2}=\left(x^{2}-a z\right)^{2}\left(a^{2}-x^{2}\right) .\)
Step-by-Step Solution
Verified Answer
Answer: The surface generated is given by the equation \(x^{4}y^{2}=(x^{2}-az)^{2}(a^{2}-x^{2})\).
1Step 1: Intersecting points with given curves
First, we will find the points where the line intersects the given curves \(x^{2}+y^{2}=a^{2}, z=0\) and \(x^{2}=az, y=0\).
For the intersection with the first curve, we have \(z=0\), so the equation becomes \(x^{2}+y^{2}=a^{2}\). Let's solve for y: \(y = \pm\sqrt{a^{2}-x^{2}}\).
For the intersection with the second curve, we have \(y=0\), so the equation becomes \(x^{2}=az\). We will solve for z: \(z= \frac{x^2}{a}\).
2Step 2: Parametric equation of the line
To obtain the parametric equation of the line, we can express it using a parameter, \(t\):
Line equation: \(x=x, y=t\sqrt{a^{2}-x^{2}}, z=t\frac{x^2}{a}\).
3Step 3: Generating the parametric surface
We can generate the parametric surface by moving the line along the x-axis while maintaining its orientation.
Parametric surface equation: \(S(x,t) = (x, t\sqrt{a^{2}-x^{2}}, t\frac{x^2}{a})\).
4Step 4: Obtaining the surface equation and comparing
Let's call the obtained parametric equations as follows:
\(y = t\sqrt{a^{2}-x^{2}}\), \(z = t\frac{x^2}{a}\).
Now, our goal is to show that this parametric surface matches the given surface equation: \(x^{4}y^{2}=(x^{2}-az)^{2}(a^{2}-x^{2})\). If we can obtain this equation from the parametric equations, we have proven that the line generates the given surface.
Let's eliminate \(t\) from the parametric equations and express \(y^{2}\) and \(z\) in terms of \(x\) and \(a\).
We have,
\(y^{2} = t^{2}(a^{2}-x^{2})\)
And,
\(t=\frac{az}{x^2}\)
Plugging t in \(y^{2}\), we get
\(y^{2} = \frac{a^{2}z^{2}(a^{2}-x^{2})}{x^{4}}\)
Multiplying both sides by \(x^{4}\), we get:
\(x^{4}y^{2}=(x^{2}-az)^{2}(a^{2}-x^{2})\)
This is the given surface equation, proving that the straight line generates the surface as claimed.
Key Concepts
Analytical Geometry3D GeometrySurface Generation
Analytical Geometry
Analytical geometry, also known as coordinate geometry, is a branch of mathematics that uses algebraic equations to represent and solve geometrical problems. It involves the study of geometric figures like points, lines, and shapes using a coordinate system, usually the Cartesian coordinate system.
Analytical geometry allows us to describe geometric shapes algebraically and understand their properties more deeply than with traditional methods alone. In our exercise, we use analytical geometry to prove that a certain line movement generates a specific surface. By finding the intersections of the straight line with given curves and expressing points on the line through a parameter, we can derive a parametric surface equation that represents the entire set of points covered by the moving line.
Analytical geometry allows us to describe geometric shapes algebraically and understand their properties more deeply than with traditional methods alone. In our exercise, we use analytical geometry to prove that a certain line movement generates a specific surface. By finding the intersections of the straight line with given curves and expressing points on the line through a parameter, we can derive a parametric surface equation that represents the entire set of points covered by the moving line.
3D Geometry
3D geometry, or three-dimensional geometry, is the study of shapes in three-dimensional space. Shapes in 3D geometry have depth, in addition to width and height, and are described in terms of length, width, and height (or depth). In our context, we deal with curves and surfaces in 3D geometry.
For instance, when we consider the curves given by the equations \(x^{2}+y^{2}=a^{2}, z=0\) and \(x^{2}=az, y=0\), we're looking at their existence within three-dimensional space. These curves and the generated surface are part of a three-dimensional Euclidean space with axes usually labeled x, y, and z. By creating a parametric equation for a line in 3D space, we are able to define a trajectory that, when followed, generates the prescribed surface.
For instance, when we consider the curves given by the equations \(x^{2}+y^{2}=a^{2}, z=0\) and \(x^{2}=az, y=0\), we're looking at their existence within three-dimensional space. These curves and the generated surface are part of a three-dimensional Euclidean space with axes usually labeled x, y, and z. By creating a parametric equation for a line in 3D space, we are able to define a trajectory that, when followed, generates the prescribed surface.
Surface Generation
Surface generation is a process in which a new surface is created through the movement of a line or curve in space. There are many methods to generate a surface, but in our exercise, we use a technique where a line is moved along the x-axis, while maintaining its orientation relative to the yz-plane.
Through this process, known as sweeping, every point on the line describes a path that collectively forms a surface. The parametric surface equation \(S(x,t) = (x, t\sqrt{a^{2}-x^{2}}, t\frac{x^2}{a})\) captures all such paths. By proving algebraically that this parametric representation conforms to the given surface equation \(x^{4}y^{2}=(x^{2}-az)^{2}(a^{2}-x^{2})\), we confirm the line's trajectory generates the desired surface in 3D space.
Through this process, known as sweeping, every point on the line describes a path that collectively forms a surface. The parametric surface equation \(S(x,t) = (x, t\sqrt{a^{2}-x^{2}}, t\frac{x^2}{a})\) captures all such paths. By proving algebraically that this parametric representation conforms to the given surface equation \(x^{4}y^{2}=(x^{2}-az)^{2}(a^{2}-x^{2})\), we confirm the line's trajectory generates the desired surface in 3D space.
Other exercises in this chapter
Problem 3
Prove that the equation of the cylinder with generators parallel to \(z\) -axis and passing through the curve \(a x^{2}+b y^{2}=2 c z, l x+m y+n z=p\) is \(n\le
View solution Problem 7
A cylinder cuts the plane \(z=0\) with curve \(x^{2}+\frac{y^{2}}{4}=\frac{1}{4}\) and has its axis parallel to \(3 x=-6 y=2 z\). Find its equation.
View solution Problem 10
Find the equation of the right circular cylinder of radius 2 whose axis passes through \((1,2,3)\) and has direction cosines proportional \(2,-3,6\).
View solution Problem 11
Find the equation of the right circular cylinder of radius 1 with axis as \(\frac{x-1}{2}=\frac{y}{3}=\frac{z-3}{1}\)
View solution