Problem 59 - A Complete Resource Book in Mathematics for JEE Main

Question

The maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with its vertex at one end of the major axis is \(\begin{array}{ll}\text { (A) } \sqrt{3} a b & \text (B) } \frac{3 \sqrt{3}}{4} a b\end{array}\) (C) \(\frac{5 \sqrt{3}}{4} a b \quad\) (D) none of these

Step-by-Step Solution

Answer

The maximum area is \(\frac{3\sqrt{3}}{4} a b\). It matches option B.

Step 1: Define Triangle Coordinates

Consider the isosceles triangle with its vertex at the endpoint of the major axis of the ellipse. Without loss of generality, place the vertex at coordinate \((a, 0)\). The other two vertices of the triangle will symmetrically lie on the ellipse, say at coordinates \((x, y)\) and \((x, -y)\). This ensures the base of the triangle is perpendicular to the major axis.

Step 2: Apply Ellipse Equation Constraint

Since the points \((x, y)\) and \((x, -y)\) lie on the ellipse, they must satisfy the ellipse equation: \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.\]Use this constraint as we proceed to maximize the area.

Step 3: Derive Area Formula of Triangle

The area \(A\) of the triangle can be expressed using the formula for the area of a triangle given base and height: \[A = \frac{1}{2} \times \text{base} \times \text{height}.\] For the triangle, the base is the distance between \((x, y)\) and \((x, -y)\), which is \(2y\), and the height is the distance from this base to the opposite vertex \((a, 0)\), which is \((a - x)\). Thus: \[ A = \frac{1}{2} \times 2y \times (a - x) = y(a - x). \]

Step 4: Substitute and Simplify Using Constraints

Using the equation of the ellipse, express \(y\) in terms of \(x\): \[y = b \sqrt{1 - \frac{x^2}{a^2}}.\]Substitute this into the area expression, \[A = b \sqrt{1 - \frac{x^2}{a^2}} (a - x).\]

Step 5: Optimize Area Expression

To find the maximum area, differentiate the expression for \(A\) with respect to \(x\), set the derivative equal to zero, and solve for \(x\). Utilizing the product rule and the chain rule, carry out the differentiation to obtain the expression for the derivative.

Step 6: Solve for Local Extrema

After differentiating, set \(\frac{dA}{dx} = 0\) to find critical points. Solve for \(x\), checking valid solutions that lie within the domain defined by the ellipse equation, \( |x| \leq a \). This gives you potential extrema for area.

Step 7: Evaluate Maximum Area

Substitute the value of \(x\) from the critical points back into the area function to find the maximum possible area. Ensure the value satisfies the condition for extremum (either directly or by comparison) to confirm if it provides the maximum value.

Key Concepts

Isosceles Triangle

An isosceles triangle is a type of triangle with at least two equal sides. In the context of our problem, this triangle is inscribed within an ellipse. It has its vertex at the end of the major axis, making it symmetrically positioned around the major axis of the ellipse.

• The vertex of our isosceles triangle is placed at the point \((a, 0)\), on the major axis.
• The other two vertices lie at points \((x, y)\) and \((x, -y)\) on the ellipse.
• These points create a symmetry with respect to the x-axis, ensuring that the base of the triangle is horizontal, thus perpendicular to the major axis.

This arrangement maximizes the potential base of the triangle, facilitating the quest to find a triangle with the largest possible area within these geometric constraints.

Major Axis

The major axis of an ellipse is the longest diameter that runs through its center. It's an important structural feature that defines the overall shape of the ellipse.
In this context:

• The major axis extends from one side of the ellipse to the other, passing through the center.
• Its endpoints on the x-axis are \( (a, 0) \) and \( (-a, 0) \), where \(a\) is the semi-major axis length.
• In this problem, the vertex of our isosceles triangle is precisely at \( (a, 0) \), precisely at one end of the major axis.

This alignment allows the triangle to leverage the ellipse's longest axis, thus providing potential for the maximum area.

Maximum Area

Finding the maximum area of the triangle involves utilizing calculus and geometric principles. Here’s a simple breakdown of the steps:

• The base of the triangle is \(2y\), derived from the coordinates \((x, y)\) and \((x, -y)\).
• The height is \(a - x\), measured from the base to the opposite vertex \((a, 0)\).
• The area formula is \(A = y(a - x)\).
• To maximize the area, express \(y\) using the ellipse equation and substitute it back, resulting in \(A = b \sqrt{1 - \frac{x^2}{a^2}} (a - x)\).
• Maximize \(A\) by differentiating with respect to \(x\) and finding critical points.

This approach uses calculus to explore valid values of \(x\) that deliver the greatest possible area for this inscribed isosceles triangle.

Critical Points

Critical points in calculus are values where the function's derivative is zero or undefined. They are potential locations for maxima or minima.
In the context of this problem:

• Differentiating the area function \(A(x) = b \sqrt{1 - \frac{x^2}{a^2}} (a - x)\) identifies where changes in the rate of the area occur.
• Set the derivative \(\frac{dA}{dx} = 0\) to find these critical points.
• Solve for \(x\) within the bounds \([-a, a]\) due to our ellipse constraints.

Evaluating these critical points confirms which provides the peak area - the target of the entire exercise. This process highlights the magic of calculus in optimizing such conditional geometries.