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Question 94 of 126

Q.Find the equation and eccentricity of the ellipse if:

(i) the centre of ellipse is same as centre of the hyperbola 4x2−9y2+8x−36y−68=04x^2 - 9y^2 + 8x - 36y - 68 = 0
(ii) length of semi major axis is 3
(iii) length of minor axis is 2
(iv) the equation of major axis is x=−1x = -1
Puducherry TnboardTamil Nadu HSC (DGE) Board 2018Subjective· 10mImportance★★★★★
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The required ellipse has centre (−1,−2)(-1,-2) (the hyperbola's centre), semi-axes a=3, b=1a=3,\ b=1 with a vertical major axis x=−1x=-1, giving eccentricity e=223e=\dfrac{2\sqrt2}{3}.

  1. Find the centre of the hyperbola 4x2−9y2+8x−36y−68=04x^2-9y^2+8x-36y-68=0 by completing the square: 4(x2+2x)−9(y2+4y)=684(x^2+2x) - 9(y^2+4y) = 68 4(x2+2x+1)−4−9(y2+4y+4)+36=684(x^2+2x+1) - 4 - 9(y^2+4y+4) + 36 = 68 4(x+1)2−9(y+2)2=364(x+1)^2 - 9(y+2)^2 = 36.
  2. So the centre of the hyperbola is (−1,−2)(-1,-2); by the given condition (i), this is also the centre of the required ellipse.
  3. Given (ii): semi-major axis a=3a=3.
  4. Given (iii): length of minor axis =2=2, so semi-minor axis b=1b=1.
  5. Given (iv): the major axis is the line x=−1x=-1, a vertical line through the centre (−1,−2)(-1,-2) — so the major axis is parallel to the yy-axis (i.e. the ellipse is 'taller' than wide). …

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