Question

Let S and S' be the foci of the ellipse and B be any one of the extremities of its minor axis. If 'S'BS is a right angled triangle with right angle at B and area ($\Delta$S'BS) = 8 s units, then the length of a latus rectum of the ellipse is :

• A. $2\sqrt{2}$
• B. 2
• C. 4
• D. 4$\sqrt{2}$

Answer 1

The Correct Option is C

To solve this problem, we need to make use of the properties of an ellipse and the given conditions involving the triangle formed by the foci and the extremity of the minor axis.

Let's start by understanding and defining some important parameters of the ellipse:

• The foci of the ellipse are generally represented as \( S(c, 0) \) and \( S'(-c, 0) \).
• The endpoints of the minor axis of the ellipse are \( (0, b) \) and \( (0, -b) \). We consider \( B(0, b) \) as one extremity of the minor axis.

Given that:

• The triangle \( S'BS \) is a right-angle triangle with a right angle at \( B \).
• The area of \(\Delta S'BS\) is 8 square units.

Using the formula for the area of a triangle with vertices \( A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) \), the area is given by:

\(Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|\)

For triangle \( S'BS \), the coordinates are \( S'(-c, 0), B(0, b), S(c, 0) \). Using these coordinates, we calculate the area:

\(\frac{1}{2} \left| -c(b - 0) + 0(0 - 0) + c(0 - b) \right| = \frac{1}{2} \left| -cb - cb \right| = cb\)

We are given that the area is 8 square units:

\(cb = 8\) (Equation 1)

Since \( S'BS \) is a right triangle with the right angle at \( B \), by Pythagoras' theorem, we have:

\(SS'^2 = SB^2 + S'B^2\), which simplifies to:

\((2c)^2 = (c^2 + b^2) + (c^2 + b^2) = 2(c^2 + b^2)\)

This implies:

\(4c^2 = 2(c^2 + b^2)\)

\(2c^2 = c^2 + b^2\)

\(c^2 = b^2 \Rightarrow b = c\)

Substituting \( b = c \) in Equation 1:

\(c^2 = 8 \Rightarrow c = \sqrt{8} = 2\sqrt{2}\)

In the context of an ellipse, the semi-major axis \( a \), semi-minor axis \( b \), and the distance to the focus \( c \) are related by:

\(a^2 = b^2 + c^2 = 8 + 8 = 16 \Rightarrow a = 4\)

The length of the latus rectum \( l \) of the ellipse is given by:

\(l = \frac{2b^2}{a} = \frac{2 \times 8}{4} = 4\)

Therefore, the length of the latus rectum of the ellipse is \(4\).

Hence, the correct answer is: 4.