Problem 99 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Given: A circle, \(2 x^{2}+2 y^{2}=5\) and a parabola, \(y^{2}=4 \sqrt{5 x}\) Statement-I: An equation of a common tangent to these curves is \(y=x+\sqrt{5}\). Statement - II: If the line, \(y=m x+\frac{\sqrt{5}}{m}(m \neq 0)\) is their common tangent, then \(m\) satisfies \(m^{4}-3 m^{2}+2=0\) (A) Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I (B) Statement -I is True; Statement -II is False. (C) Statement -I is False; Statement -II is True (D) Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I

Step-by-Step Solution

Answer

Statement-I is False; Statement-II is True. (C)

Step 1: Analyze the Circle Equation

The given circle is \(2x^2 + 2y^2 = 5\). Divide by 2 to get it in standard form: \(x^2 + y^2 = \frac{5}{2}\). This represents a circle centered at the origin \((0, 0)\) with radius \(\sqrt{\frac{5}{2}}\).

Step 2: Analyze the Parabola Equation

The parabola is given by \(y^2 = 4\sqrt{5x}\). This can be rewritten as \(y^2 = 4\sqrt{5} \sqrt{x}\). Since it is written as \(y^2 = 4ax\), it is a standard parabola opening to the right with vertex at \((0, 0)\) and \(4a = 4\sqrt{5}\) implying \(a = \sqrt{5}\).

Step 3: Derive the Condition for Common Tangent

A line tangent to both curves will be of the form \(y = mx + c\). We expect the tangent to be common if it satisfies the tangency condition for both curves. For the parabola, using its format, the line \(y = mx + \frac{\sqrt{5}}{m}\) should be tangent, hence satifying \(c = \frac{\sqrt{5}}{m}\) when written as \(y = mx + c\).

Step 4: Solve the Common Tangent Condition

Plugging the line \(y = mx + \frac{\sqrt{5}}{m}\) into the circle's equation and setting the discriminant to zero (for tangency), we find the equation \(m^4 - 3m^2 + 2 = 0\). Thus, this equation arises directly from the calculation ensuring it meets both the circle and parabola's tangents.

Step 5: Verify Statement I

Checking \(y = x + \sqrt{5}\) as a tangent to both curves involves verifying it satisfies both conditions derived above. However, calculating we realize the condition is not met accurately for both curves in its tangency form without previous derived method.

Step 6: Solution Reasoning

Therefore, Statement-I is false because the initial method derived does not satisfy the general tangency condition without the specific setup for \(m\), and Statement-II is true as it indeed provides the rational derivation for the common tangent.

Key Concepts

Tangency Condition

When dealing with curves, such as a circle and a parabola, identifying a common tangent involves understanding the tangency condition. A common tangent is a line that touches both curves at exactly one point each. For this to happen, the line's equation must satisfy the condition of tangency for both curves.

To determine this, the line is typically represented as a straight line equation:

• Circle: For a line to be tangent to a circle, the discriminant of the system formed by replacing the circle's equation with the line should be zero. This ensures the line intersects the circle at exactly one point.
• Parabola: For a parabola, the process can be more complex but involves substituting the line's equation into the parabola's and then ensuring only a single solution exists for the intersection.

An important takeaway is that the condition for tangency hinges on aligning these conditions simultaneously for both curves. If even a small alteration is present, such as the incorrect choice of slope or y-intercept in the tangent's equation, the line may no longer be tangent to either or both curves.

Circle Equation

The equation provided in the exercise for the circle is initially given as \(2x^2 + 2y^2 = 5\).

To analyze it effectively, divide the entire equation by 2 to bring it into its standard circle form:

• This simplifies to \(x^2 + y^2 = \frac{5}{2}\), indicating the center is at the origin, \((0,0)\).
• The radius is found by comparing with the standard form \((x-h)^2 + (y-k)^2 = r^2\), hence: - The circle's radius is \(\sqrt{\frac{5}{2}}\).

This form is critical because once we have the radius, it becomes easier to determine lines that might touch the circle exactly once. Understanding these aspects of the circle's equation is key in checking if a certain line is a tangent by focusing on its geometric properties.

Parabola Equation

The parabola in this problem is represented by the equation \(y^2 = 4\sqrt{5x}\).

This equation can be transformed into a more familiar form:

• By substituting \(4a\) with \(4\sqrt{5}\), which means \(a = \sqrt{5}\), the equation resembles \(y^2 = 4ax\).
• This is indicative of a standard parabola that opens to the right.

The principles of how

• Here, the vertex is at the origin \((0,0)\).

These details are crucial as they enable us to graph the parabola effectively and understand potential tangents. Such a visualization aids in comprehending how a line would meet the parabola at exactly one point, reflecting the tangency condition discussed earlier. Remember, the orientation and position of the parabola deeply influence potential tangent lines.