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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 x29+y25=1\dfrac{x^2}{9}+\dfrac{y^2}{5}=1 at the ends of the latus rectum. The area of the quadrilateral so formed is:
A2727correct
B272\dfrac{27}{2}
C274\dfrac{27}{4}
D2755\dfrac{27}{55}
Solution
Step 1: Here a2=9, b2=5a^2=9,\ b^2=5, so a=3, c=9−5=2, e=23a=3,\ c=\sqrt{9-5}=2,\ e=\dfrac23. The latus-rectum ends are (±2,±53)\left(\pm2,\pm\dfrac53\right). Step 2: Tangent at (2,53)\left(2,\dfrac53\right) is 2x9+y3=1\dfrac{2x}{9}+\dfrac{y}{3}=1, meeting the axes at (92,0)\left(\dfrac92,0\right) and (0,3)(0,3). By symmetry the four tangents form a rhombus with vertices (±92,0)\left(\pm\dfrac92,0\right) and (0,±3)(0,\pm3). Step 3: The diagonals are 99 and 66, so the area is
12 (9)(6)=27.\frac12\,(9)(6)=27.
 (Equivalently, area =2a2e=2⋅92/3=27=\dfrac{2a^2}{e}=\dfrac{2\cdot9}{2/3}=27.) Correct answer: (1)
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