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

**Solution:**

The given ellipse is x^{2}/a^{2} + y^{2}/b^{2} = 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 (- x_{1}, y_{1}) and the coordinates of B to be (- x_{1}, - y_{1})

Now, we have y_{1} = ± b/a √ (a^{2} - x_{1}^{2})

Coordinates of A are (- x_{1}, b/a √ (a^{2} - x_{1}^{2})

the coordinates of B are (x_{1}, - b/a √ (a^{2} - x_{1}^{2}))

As the point (x_{1}, y_{1})

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

A = | a (2b/a √ (a^{2} - x_{1}^{2})) + (- x_{1}) (- b/a √ (a^{2} - x_{1}^{2})) + (- x_{1}) (- b/a √ (a^{2} - x_{1}^{2})) |

= b √ (a^{2} - x_{1}^{2}) + x_{1} b/a √ (a^{2} - x_{1}^{2}) .... (1)

Therefore,

dA/dx_{1} = (- 2x_{1}b / 2 √ (a^{2} - x_{1}^{2})) + b / a √ (a^{2} - x_{1}^{2}) - (- 2b x_{1}^{2} / 2a √ (a^{2} - x_{1}^{2}))

= (b/a √ (a^{2} - x_{1}^{2})) [- x_{1}a + (a^{2} - x_{1}^{2}) - x_{1}^{2}]

= b(- 2x_{1}^{2} - x_{1}a + a²)/a √ (a^{2} - x_{1}^{2})

Now, dA/dx_{1} = 0

Hence,

- 2x_{1}^{2} - x_{1}a + a^{2} = 0

x_{1} = (a ± √a² - 4(- 2)(a²)) / (2(- 2))

x_{1} = (a ± √9a²) / (- 4)

x_{1} = (a ± 3a) / (- 4)

x_{1} = - a, a / 2

But x_{1} ≠ - a

Therefore,

x_{1} = a/2

y_{1} = b / a √a² - a²/4

= ba / 2a √3

= √3b / 2

Now,

d^{2}A / dx_{1}^{2} = b / a {√ (a^{2} - x_{1}^{2}) (- 4x_{1} - a) - (- 2x_{1}^{2} - x_{1}a + a^{2}) (- 2x_{1 }/ 2√ (a^{2} - x_{1}^{2})} / (a^{2} - x_{1}^{2})

= b/a {(a^{2} - x_{1}^{2})(- 4x_{1} - a) + x_{1}(- 2x_{1}^{2} - x_{1}a + a^{2})} / (a^{2} - x_{1}^{2})^{3/2}

= b/a {2x^{3} - 3a^{2}x - a^{3}} / (a^{2} -x_{1}^{2})^{3/2}

Also, when, x_{1} = a/2

Then,

d^{2}A / dx_{1}^{2} = b/a {2(a/8)^{3} - 3a^{2}(a/2) - a^{3}}/(a^{2} - (a/2)^{2})^{3/2}

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

= b/a {9/4a^{3}} / (3a^{2}/4)^{3/2} < 0

Thus, the area is the maximum when x_{1} = 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

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

**Summary:**

The maximum area of an isosceles triangle inscribed in the ellipse x^{2}/a^{2} + y^{2}/b^{2} = 1 with its vertex at one end of the major axis is (3√3 / 4) ab

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