Maths-
General
Easy

Question

Tangents are drawn to the ellipse fraction numerator x to the power of 2 end exponent over denominator 9 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 end fraction equals 1 at the ends of the latus rectum. The are of the quadrilateral so formed is

  1. 27    
  2. fraction numerator 13 over denominator 2 end fraction    
  3. fraction numerator 15 over denominator 4 end fraction    
  4. 45    

hintHint:

The eccentricity of an ellipse is always less than 1. i.e. e < 1. The eccentricity of an ellipse can be taken as the ratio of its distance from the focus and the distance from the directrix.
Eccentricity of ellipse (e) = square root of 1 minus b squared over a squared end root
Here a is the length of the semi-major axis and b is the length of the semi-minor axis.

The correct answer is: 27



    Let LL and MM are the latus rectum.
    Eccentricity of ellipse (e) = square root of 1 minus b squared over a squared end root
    Here a is the length of the semi-major axis and b is the length of the semi-minor axis.
    e space equals space square root of 1 space minus space b squared over a squared end root
F r o m comma space x squared over 9 space plus space y squared over 5 space equals 1
b squared space equals space 5 space a n d space a squared space equals space 9

e space equals space square root of 1 space minus space 5 over 9 end root
e space equals space 2 over 3
    The equation of tangent at (x1,y1) is fraction numerator x x subscript 1 over denominator 9 end fraction space plus space fraction numerator y y subscript 1 over denominator 5 end fraction space equals space 1
    The tangent passes through point L (ae, b squared over a right parenthesis
    S o comma space fraction numerator x left parenthesis a e right parenthesis over denominator 9 end fraction space plus space fraction numerator y left parenthesis begin display style b squared over a end style right parenthesis over denominator 5 end fraction space equals 1
space space space space space space rightwards double arrow e x space plus space y space equals space a
space space space space space space rightwards double arrow fraction numerator x over denominator begin display style a over e end style end fraction plus space y over a space equals space 1
    So, the tangent cuts the x-axis at point A(a/e,0) and y-axis at point B(0,a)
    Area of quadrilateral ABC=  4×area of OAB
    Area of quadrilateral ABCD =   4 space cross times 1 half cross times a over e cross times a
    Area of quadrilateral ABCD = 2 space cross times a squared over e
    Area of quadrilateral ABCD  = 2 space space cross times space fraction numerator 9 over denominator begin display style 2 over 3 end style end fraction
    Area of quadrilateral ABCD  = 27sq. units

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