Problem 115 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Equation of a common tangent to the circle, \(x^{2}+y^{2}-6 x=0\) and the parabola, \(y^{2}=4 x\), is : (a) \(2 \sqrt{3} y=12 x+1\) (b) \(\sqrt{3} y=x+3\) (c) \(2 \sqrt{3} y=-x-12\) (d) \(\sqrt{3} y=3 x+1\)

Step-by-Step Solution

Answer

The correct equation of the common tangent is (d) \(\sqrt{3} y = 3x + 1\).

Step 1: Identify Equations of Given Figures

We are given two equations:1. The circle: \(x^2 + y^2 - 6x = 0\)2. The parabola: \(y^2 = 4x\)First, rearrange the circle's equation into standard form. Completing the square for \(x^2 - 6x\), we get:\[(x-3)^2 + y^2 = 9\]This represents a circle centered at (3, 0) with a radius of 3. The parabola is already in its standard form, opening to the right at vertex (0,0).

Step 2: Determine Tangent Line Form

The equation of a tangent line to a circle or a parabola can be expressed in the form:\[ y = mx + c \]For a tangent line to both the circle and the parabola, the tangent must touch both curves without crossing them. We suppose a line \(y = mx + c\) as common tangent to both figures.

Step 3: Condition for Tangency to the Circle

Substitute \(y = mx + c\) into the circle's equation \((x - 3)^2 + y^2 = 9\):\[ (x - 3)^2 + (mx + c)^2 = 9 \]Expanding and simplifying will give a quadratic in \(x\). For tangency, this quadratic should have exactly one solution (discriminant = 0). Solving for this condition gives an equation in terms of \(m\) and \(c\).

Step 4: Condition for Tangency to the Parabola

Substitute \(y = mx + c\) into the parabola's equation \(y^2 = 4x\):\( (mx + c)^2 = 4x \)Expanding gives a quadratic equation in \(x\). For this to be a tangent line, the discriminant of this quadratic must also be zero. This will provide us another equation in terms of \(m\) and \(c\).

Step 5: Solve for m and c

Using the conditions derived from both tangency conditions, solve the equations simultaneously to find the values of \(m\) and \(c\) that allow the line to be tangent to both the circle and the parabola. Solving typically involves determining \(m\) first from the circle's condition and then substituting back to find \(c\).

Step 6: Identify the Correct Tangent Equation

Insert the obtained values of \(m\) and \(c\) back into the tangent line equation \(y = mx + c\). Compare this equation with the given options (a-d). The goal is to match one of these option equations with the derived equation by verifying both the slope and intercept.

Key Concepts

Geometry Equations

Geometry involves various shapes and figures that have specific equations to represent their properties. When it comes to understanding equations in geometry, we often encounter different types of conic sections, such as circles, ellipses, hyperbolas, and parabolas.
Each of these figures has a specific equation that describes its shape and position in the plane.

• A circle's equation typically involves both the variables \(x\) and \(y\), capturing its symmetry around the center point.
• A parabola's equation is more linear, typically involving one squared term, which represents its open-end direction.
• Total comprehension of these equations requires completing squares in circles and aligning terms in parabolas to standard forms.

These equations are crucial in solving problems involving intersection points and tangents between different geometric figures.

Circle Equation

The circle is one of the simplest forms of the conic sections, and its equation can reveal much about the circle's properties.
For this particular problem, we are given an equation of the circle: \[ x^2 + y^2 - 6x = 0 \]
To express this more informatively, we complete the square.

• In the term \(x^2 - 6x\), add \(9\) to both sides to complete the square.
• The equation converts to \((x - 3)^2 + y^2 = 9\).
• This reveals that the circle is centered at \((3, 0)\) with a radius of \(3\).

Completing the square is an essential skill as it simplifies the equation, illustrating key features like the center and radius, which are necessary for analyzing tangents.

Parabola Equation

Parabolas have a distinct feature in their ability to open either horizontally or vertically, different from the symmetry of circles. In our case, the given parabola equation is:\[ y^2 = 4x \]
This equation reveals several characteristics about the parabola.

• The parabola opens horizontally because the \(y\) term is squared.
• It is centered at the origin \((0, 0)\).
• The coefficient \(4\) in front of \(x\) gives information about the width and direction of the opening.

Understanding the parabola’s equation helps us in determining lines such as tangents, which only touch the parabola at one point without crossing it.

Tangent Condition

A tangent is a line that just touches the curve of a circle or a parabola at precisely one point without intersecting it. The condition for a line to be tangent involves examining the discriminant of equations derived from substituting the line equation.

• For the circle's equation \((x-3)^2 + (mx+c)^2 = 9\), expand the terms to form a quadratic expression. The discriminant must be zero to suggest tangency.
• For the parabola \((mx + c)^2 = 4x\), also expand to a quadratic form and set the discriminant to zero.
• In both cases, the condition of zero discriminant ensures a single solution, indicating the point of tangency.

By fulfilling these conditions, we can find the exact parameters for our line (slope \(m\) and intercept \(c\)) making it tangent to both the circle and the parabola simultaneously.