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Ellipse question

2017 · 8 Apr · Shift 1 · Q34
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Ellipse question

2017 · 8 Apr · Shift 1 · Q34

JEE MainMathematicsEllipseMCQ+4 / −1
Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is 35{3 \over 5}53​ and the distance between its foci is 6, then the area (in sq. units) of the quadrilatateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is :
  1. A
    8
  2. B
    32
  3. C
    80
  4. D
    40
View written solutionFree

Correct answer: D

  1. Write the standard form of the ellipse

Since the center is at the origin and the major axis is along the xxx-axis, the ellipse is of the form

x2a2+y2b2=1,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a2x2​+b2y2​=1,

where a>ba>ba>b, eccentricity e=cae=\frac{c}{a}e=ac​, and foci are at (±c,0)(\pm c,0)(±c,0).

  1. Use the given distance between foci

Distance between the foci is 666, so

2c=6  ⟹  c=3.2c=6 \implies c=3.2c=6⟹c=3.
  1. Use the eccentricity

Given

e=35=ca.e=\frac{3}{5}=\frac{c}{a}.e=53​=ac​.

Substituting c=3c=3c=3,

35=3a  ⟹  a=5.\frac{3}{5}=\frac{3}{a} \implies a=5.53​=a3​⟹a=5.
  1. Find bbb using the ellipse relation

For an ellipse,

c2=a2−b2.c^2=a^2-b^2.c2=a2−b2.

So,

b2=a2−c2=25−9=16  ⟹  b=4.b^2=a^2-c^2=25-9=16 \implies b=4.b2=a2−c2=25−9=16⟹b=4.
  1. Identify the vertices of the ellipse

The four vertices are

(±a,0)=(±5,0),(0,±b)=(0,±4).(\pm a,0)=(\pm 5,0), \qquad (0,\pm b)=(0,\pm 4).(±a,0)=(±5,0),(0,±b)=(0,±4).

These four points form a quadrilateral (actually a rhombus).

  1. Find the area of the quadrilateral

Its diagonals are:

  • horizontal diagonal =2a=10=2a=10=2a=10
  • vertical diagonal =2b=8=2b=8=2b=8

Area of a quadrilateral with perpendicular diagonals is

12×d1×d2.\frac{1}{2}\times d_1 \times d_2.21​×d1​×d2​.

Hence,

Area=12×10×8=40.\text{Area}=\frac{1}{2}\times 10 \times 8=40.Area=21​×10×8=40.
  1. Match with the options
40\boxed{40}40​

So the correct option is D.

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