Find the maximum area of an isosceles triangle inscribed in the ellipse x²/a² + y²/b² = 1 with its vertex at one end of the major axis

Solution:

Find the maximum area of an isosceles triangle inscribed in the ellipse

The given ellipse is x²/a² + y²/b² = 1

Let the major axis be along the x-axis.

Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0) .

Since the ellipse is symmetrical with respect to the x-axis and y-axis, we can assume the coordinates of A to be (- x₁, y₁) and the coordinates of B to be (- x₁, - y₁)

Now, we have y₁ = ± b/a √ (a² - x₁²)

Coordinates of A are (- x₁, b/a √ (a² - x₁²)

the coordinates of B are (x₁, - b/a √ (a² - x₁²))

As the point (x₁, y₁)

lies on the ellipse, the area of triangle ABC (A) is given by,

A = | a (2b/a √ (a² - x₁²)) + (- x₁) (- b/a √ (a² - x₁²)) + (- x₁) (- b/a √ (a² - x₁²)) |

= b √ (a² - x₁²) + x₁ b/a √ (a² - x₁²) .... (1)

Therefore,

dA/dx₁ = (- 2x₁b / 2 √ (a² - x₁²)) + b / a √ (a² - x₁²) - (- 2b x₁² / 2a √ (a² - x₁²))

= (b/a √ (a² - x₁²)) [- x₁a + (a² - x₁²) - x₁²]

= b(- 2x₁² - x₁a + a²)/a √ (a² - x₁²)

Now, dA/dx₁ = 0

Hence,

- 2x₁² - x₁a + a² = 0

x₁ = (a ± √a² - 4(- 2)(a²)) / (2(- 2))

x₁ = (a ± √9a²) / (- 4)

x₁ = (a ± 3a) / (- 4)

x₁ = - a, a / 2

But x₁ ≠ - a

Therefore,

x₁ = a/2

y₁ = b / a √a² - a²/4

= ba / 2a √3

= √3b / 2

Now,

d²A / dx₁² = b / a {√ (a² - x₁²) (- 4x₁ - a) - (- 2x₁² - x₁a + a²) (- 2x₁/ 2√ (a² - x₁²)} / (a² - x₁²)

= b/a {(a² - x₁²)(- 4x₁ - a) + x₁(- 2x₁² - x₁a + a²)} / (a² - x₁²)^3/2

= b/a {2x³ - 3a²x - a³} / (a² -x₁²)^3/2

Also, when, x₁ = a/2

Then,

d²A / dx₁² = b/a {2(a/8)³ - 3a²(a/2) - a³}/(a² - (a/2)²)^3/2

= b/a {a³/4 - 3/2a³ - a³}/(a² - (a/2)²)^3/2

= b/a {9/4a³} / (3a²/4)^3/2 < 0

Thus, the area is the maximum when x₁ = a / 2

Hence, Maximum area of the triangle is given by,

A = b √a² - a²/4 + (a/2)b/a √a² - a²/4

= ab √3 / 2 + ab √ 3/4

= (3√3 / 4) ab

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 8