Problem 108 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

Option (A) is correct because the curves cut orthogonally for \(c=0\), which is in the range \((0,16)\).

Step 1: 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 \(x^2 - 16x + c = 0\).

Step 2: 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 \(c\).

Step 3: 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 \(\frac{x}{4y} \times \frac{2}{y} = -1\).

Step 4: 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 \(x = 0\).

Step 5: 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.

Key Concepts

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 \( y \).

• 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 \( c \):

• Here, \( a = 1 \), \( b = -16 \), and \( c \) is kept as a variable.
• Substitute these values back into the quadratic formula.
• The formula results in \( x = \frac{16 \pm \sqrt{256 - 4c}}{2} \).

This equation helps determine the \( x \)-coordinates of the intersection points between the curves. At these points, we use further conditions to check for orthogonality.

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.

• Use substitution to find common intersection points.
• Solve the resulting equation to determine \( x \).
• Apply the orthogonality condition to find valid points and parameters.

Understanding intersections gives insights into the underlying structure of the geometric shapes, determining crucial points like tangents and normals while mapping out their relative positioning.