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Area of Quadrilateral from Tangents at Latus Rectum Ends | JEE

JEE Maths question with a full step-by-step solution.

Question

Tangents are drawn to the ellipse

\dfrac{x^2}{9}+\dfrac{y^2}{5}=1

at the ends of the latus rectum. The area of the quadrilateral so formed is:

27

correct

\dfrac{27}{2}

\dfrac{27}{4}

\dfrac{27}{55}

Solution

Step 1: Here

a^2=9,\ b^2=5

, so

a=3,\ c=\sqrt{9-5}=2,\ e=\dfrac23

. The latus-rectum ends are

\left(\pm2,\pm\dfrac53\right)

. Step 2: Tangent at

\left(2,\dfrac53\right)

\dfrac{2x}{9}+\dfrac{y}{3}=1

, meeting the axes at

\left(\dfrac92,0\right)

and

(0,3)

. By symmetry the four tangents form a rhombus with vertices

\left(\pm\dfrac92,0\right)

and

(0,\pm3)

. Step 3: The diagonals are

9

and

6

, so the area is

\frac12\,(9)(6)=27.

(Equivalently, area

=\dfrac{2a^2}{e}=\dfrac{2\cdot9}{2/3}=27

.) Correct answer: (1)

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