Problem 94 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the common tangent to the parabolas, \(y^{2}=4 x\) and \(x^{2}=4 y\) also touches the circle, \(x^{2}+y^{2}=\mathrm{c}^{2}\), then \(\mathrm{c}\) is equal to: (a) \(\frac{1}{2 \sqrt{2}}\) (b) \(\frac{1}{\sqrt{2}}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The value of \( c \) is \( \frac{1}{\sqrt{2}} \) (option b).

Step 1: Identify Parabola Equations

The given parabolas are:1. \( y^2 = 4x \) which opens to the right.2. \( x^2 = 4y \) which opens upwards.

Step 2: Find the Tangent Equation

Since a common tangent exists for the two parabolas, we find the tangent equations:- For \( y^2 = 4x \), the tangent in the form \( y = mx + \frac{1}{m} \) (since \( c = \frac{1}{m} \) here).- For \( x^2 = 4y \), the tangent equation is \( x = my + \frac{1}{m} \).Equating and solving gives the slope as \( m = 1 \). So one common tangent equation is \( y = x + 1 \).

Step 3: Apply Circle Condition

The line \( y = x + 1 \) should also be tangent to the circle \( x^2 + y^2 = c^2 \). Substituting \( y = x + 1 \) in the circle's equation:\( x^2 + (x + 1)^2 = c^2 \) \( 2x^2 + 2x + 1 = c^2 \).Since it is a tangent, the discriminant should be zero, giving a condition for \( c \).

Step 4: Solve for c

The discriminant of the quadratic equation in \( x \) formed is zero when touching gives a specific condition:The quadratic is \( 2x^2 + 2x + (1 - c^2) = 0 \).The discriminant (\( b^2 - 4ac \)) for this is \( 2^2 - 4(2)(1 - c^2) = 0 \).Solving gives:\( 4 - 8 + 8c^2 = 0 \)\( 8c^2 = 4 \)\( c^2 = \frac{1}{2} \)Hence, \( c = \frac{1}{\sqrt{2}} \).

Step 5: Final Answer

Therefore, \( c \) such that the common tangent also touches the circle is \( c = \frac{1}{\sqrt{2}} \), option (b).

Key Concepts

Parabolas

When we talk about parabolas in mathematics, we're referring to a type of curve that can take various shapes based on its equation. In this exercise, there are two specific parabolas considered:

• The equation \( y^2 = 4x \) represents a parabola that opens to the right. This is because the square term is associated with \( y \), and \( x \) is the independent variable. The coefficient of 4 means that this parabola is moderately wide.
• On the other hand, \( x^2 = 4y \) represents a parabola opening upwards, with \( y \) being the dependent variable and \( x \) the independent variable. The structure is similar to the first, mirroring its symmetry but in a vertical direction.

Both these parabolas exhibit a symmetry through their axes. The one opening right has a line symmetry along the x-axis, whereas the one opening upwards has its axis of symmetry along the y-axis.
The common element in their equations is the square term being set equal to 4 times a variable, influencing their shape and opening direction.

Common Tangent

A common tangent is a single line that touches two curves at different points. It's like a thread that gently touches both sides of different bowls. In our exercise, the question is about finding such a tangent line for the parabolas given by \( y^2 = 4x \) and \( x^2 = 4y \).
To find the common tangent, we explore the general tangent forms for each parabola:

• For \( y^2 = 4x \), the tangent line can be expressed as \( y = mx + \frac{1}{m} \). This is derived from transforming the parabola into its slope-intercept form.
• For \( x^2 = 4y \), its tangent has the form \( x = my + \frac{1}{m} \). This resembles the previous equation, but swaps the roles of \( x \) and \( y \) reflecting its upward opening.

To determine the common tangent, equate these equations and solve for \( m \), known as the slope. In this particular problem, \( m = 1 \) was derived, indicating the tangent line equation is \( y = x + 1 \). This line touches both parabolas effectively.

Circle Tangent Property

Imagine a small circle gently touching a line at just one point; that's the property of a tangent line to a circle. In the exercise, the problem adds a circle into the mix, given by \( x^2 + y^2 = c^2 \). The task is to ensure that the previously found common tangent \( y = x + 1 \) also tangentially touches this circle.
This means substituting \( y = x + 1 \) into the circle's equation and finding that \( c \, \)-value makes this line a true tangent. As such, the process involves plugging in \( y = x + 1 \) into the circle equation resulting in:
\[ x^2 + (x+1)^2 = c^2 \]
Simplifying gives \( 2x^2 + 2x + 1 = c^2 \). For the line to be a tangent, the quadratic equation for \( x \) must satisfy certain conditions. The specific condition here is that the discriminant must be zero, indicating a single unique solution at the tangent point.
Solving for \( c \), it was found that \( c = \frac{1}{\sqrt{2}} \). This solution confirms that the common tangent touches the circle exactly once, affirming its tangent property.