Question 71 of 82advanced🔧 ApplyLong Answer6 marks

Find the maximum area of an isosceles triangle inscribed in the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 with its vertex at one end of the major axis.

Correct Answer

The maximum area of the isosceles triangle is 334 ab\frac{3\sqrt{3}}{4}\, ab.

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 5 | (Page 38)
For More Understanding

Explanation

This is an optimization problem from the Application of Derivatives chapter. The ellipse has its major axis along the x-axis with endpoints at (±a,0)(\pm a, 0).

Taking the vertex at (a,0)(a, 0), the other two vertices of the isosceles triangle must be symmetric about the x-axis at (x,y)(x, y) and (x,−y)(x, -y).

Using the ellipse equation to express yy in terms of xx, then finding the area function and differentiating to find critical points, we determine the maximum area.

Solution Steps

  1. Step 1: Let the vertex of the isosceles triangle be at (a,0)(a, 0), one end of the major axis. Since the triangle is isosceles with vertex on the x-axis, the other two vertices must be symmetric about the x-axis. Let them be at (x,y)(x, y) and (x,−y)(x, -y) where x<ax < a.

  2. Step 2: Since these points lie on the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, we get:

    y=baa2−x2y = \frac{b}{a}\sqrt{a^2 - x^2}

  3. Step 3: The base of the triangle has length 2y2y, and the height is (a−x)(a - x). Therefore, Area:

    A=12×2y×(a−x)=y(a−x)A = \frac{1}{2} \times 2y \times (a - x) = y(a - x)

    Substituting yy:

    A=ba(a−x)a2−x2A = \frac{b}{a}(a - x)\sqrt{a^2 - x^2}

  4. Step 4: Differentiate AA with respect to xx using the product rule:

    dAdx=ba[−a2−x2+(a−x)⋅−xa2−x2]=ba⋅−(a2+ax−2x2)a2−x2\frac{dA}{dx} = \frac{b}{a}\left[-\sqrt{a^2 - x^2} + (a - x) \cdot \frac{-x}{\sqrt{a^2 - x^2}}\right] = \frac{b}{a} \cdot \frac{-(a^2 + ax - 2x^2)}{\sqrt{a^2 - x^2}}

  5. Step 5: Set dAdx=0\frac{dA}{dx} = 0:

    a2+ax−2x2=0a^2 + ax - 2x^2 = 0

    which gives:

    2x2−ax−a2=02x^2 - ax - a^2 = 0

    Solving:

    x=a±3a4x = \frac{a \pm 3a}{4}

    so x=ax = a or x=−a2x = -\frac{a}{2}.

  6. Step 6: x=ax = a gives A=0A = 0 (minimum). At x=−a2x = -\frac{a}{2}:

    y=b32y = \frac{b\sqrt{3}}{2}

    and:

    A=b32×3a2=334 abA = \frac{b\sqrt{3}}{2} \times \frac{3a}{2} = \frac{3\sqrt{3}}{4}\, ab

  7. Step 7: Verify maximum using second derivative or note that A→0A \to 0 at boundaries. The second derivative is negative at x=−a2x = -\frac{a}{2}, confirming maximum area =334 ab= \frac{3\sqrt{3}}{4}\, ab.

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