Chapter 14: Problem 108
The curves \(x^{2}-4 y^{2}+c=0\) and \(y^{2}=4 x\) will cut orthogonally for (A) \(c \in(0,16)\) (B) \(c \in(-3,4)\) (C) \(c \in(3,4)\) (D) None of these
Short Answer
Expert verified
Option (A) is correct because the curves cut orthogonally for \(c=0\), which is in the range \((0,16)\).
Step by step solution
01
Intersection of the Curves
The first curve is given by \(x^2 - 4y^2 + c = 0\), which is a hyperbola. The second curve is \(y^2 = 4x\), which is a parabola. The intersection points are found by solving these equations together. Substitute \(y^2 = 4x\) into \(x^2 - 4y^2 + c = 0\): \[ x^2 - 4(4x) + c = 0 \] This simplifies to the quadratic equation \(x^2 - 16x + c = 0\).
02
Solving for Intersection Points
To find the x-coordinates of the intersection points, solve the quadratic equation \(x^2 - 16x + c = 0\). Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -16, c = c\). This yields the roots: \[ x = \frac{16 \pm \sqrt{256 - 4c}}{2} \] Thus, the intersections depend on the value of \(c\).
03
Condition for Orthogonal Intersection
For two curves to intersect orthogonally, their gradients at the intersection points must multiply to \(-1\). Compute the derivatives:- For \(x^2 - 4y^2 + c = 0\): Use implicit differentiation to get \(2x - 8yy' = 0\), so \(y' = \frac{x}{4y}\).- For \(y^2 = 4x\): Differentiate to get \(2y dy/dx = 4\), so \(y' = \frac{2}{y}\).At the intersection points, the product should be \(\frac{x}{4y} \times \frac{2}{y} = -1\).
04
Setting the Orthogonality Condition
From \(\frac{x}{4y} \times \frac{2}{y} = -1\), simplify to: \[ \frac{2x}{4y^2} = -1 \rightarrow \frac{x}{2y^2} = -1 \rightarrow x = -2y^2 \]Substituting \(y^2 = 4x\) into \(x = -2y^2\) gives \(x = -8x\), leading to \(9x=0\), which can be solved if \(x = 0\).
05
Checking Values of 'c' for Orthogonal Intersection
Returning to the equation from the intersection point condition, substitute \(x = 0\) into the simplified quadratic equation: From \(x^2 - 16x + c = 0\) at \(x=0\), we have \(c=0\). Thus, the curves cut orthogonally when \(c = 0\). Evaluate the given options for this condition.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Implicit Differentiation
Implicit differentiation is a technique used when dealing with equations where one variable is not isolated on one side. This is common in equations involving two or more variables in a relational context, rather than functional. For example, if we differentiate an equation such as \( x^2 - 4y^2 + c = 0 \), we cannot solve directly for \( y \) because \( y \) is not expressed as a function of \( x \). Instead, we differentiate both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \).
The original equation \( x^2 - 4y^2 + c = 0 \) becomes \( 2x - 8yy' = 0 \) after differentiating, where \( y' = \frac{dy}{dx} \) is the derivative of \( y \) with respect to \( x \). Solving for \( y' \) gives \( y' = \frac{x}{4y} \). This allows us to find the slope of the curve at any given point on the curve by substituting the respective values of \( x \) and \( y \).
The original equation \( x^2 - 4y^2 + c = 0 \) becomes \( 2x - 8yy' = 0 \) after differentiating, where \( y' = \frac{dy}{dx} \) is the derivative of \( y \) with respect to \( x \). Solving for \( y' \) gives \( y' = \frac{x}{4y} \). This allows us to find the slope of the curve at any given point on the curve by substituting the respective values of \( x \) and \( y \).
- Start by differentiating each term separately.
- Remember chain rule: \( 2y \cdot \frac{dy}{dx} = 4 \).
- Solve for \( \frac{dy}{dx} \) to get the slope.
Quadratic Equation
A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \). The solutions to such equations can be found using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In the context of our problem, the quadratic equation \( x^2 - 16x + c = 0 \) was derived from intersecting the hyperbola and the parabola equations.
Solving this quadratic equation using the quadratic formula means identifying the values of \( a \), \( b \), and \( c \):
Solving this quadratic equation using the quadratic formula means identifying the values of \( a \), \( b \), and \( c \):
- Here, \( a = 1 \), \( b = -16 \), and \( c \) is kept as a variable.
- Substitute these values back into the quadratic formula.
- The formula results in \( x = \frac{16 \pm \sqrt{256 - 4c}}{2} \).
Intersection of Curves
Intersection of curves refers to the points where two or more curves meet. In this exercise, we analyze the intersection points between a hyperbola and a parabola given by the equations \( x^2 - 4y^2 + c = 0 \) and \( y^2 = 4x \). Intersection points are crucial for understanding how curves relate in their geometric space and are often found by solving equations simultaneously.
Substituting the parabola's equation \( y^2 = 4x \) into the hyperbola's equation gives us a single quadratic equation \( x^2 - 16x + c = 0 \). By solving this equation, we find the \( x \)-coordinates of the intersection points.
To find where these curves intersect orthogonally, we check the condition where the slopes of the curves at the intersection point multiply to \(-1\). This involves using the derivatives of each curve, found through implicit differentiation, to compute the slopes at these points.
Substituting the parabola's equation \( y^2 = 4x \) into the hyperbola's equation gives us a single quadratic equation \( x^2 - 16x + c = 0 \). By solving this equation, we find the \( x \)-coordinates of the intersection points.
To find where these curves intersect orthogonally, we check the condition where the slopes of the curves at the intersection point multiply to \(-1\). This involves using the derivatives of each curve, found through implicit differentiation, to compute the slopes at these points.
- Use substitution to find common intersection points.
- Solve the resulting equation to determine \( x \).
- Apply the orthogonality condition to find valid points and parameters.