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NCERT Exemplar · Q41

Q.The length of the latus rectum of the ellipse 3x2+y2=123x^2 + y^2 = 12 is
(A) 44
(B) 33
(C) 88
(D) 43\dfrac{4}{\sqrt{3}}

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Writing 3x2+y2=123x^2 + y^2 = 12 as x24+y212=1\frac{x^2}{4} + \frac{y^2}{12} = 1 gives a2=12a^2 = 12, b2=4b^2 = 4, so the latus rectum 2b2a=43\frac{2b^2}{a} = \frac{4}{\sqrt{3}} — option (D).

1. Convert to standard form. Divide 3x2+y2=123x^2 + y^2 = 12 by 1212:

3x212+y212=1  ⇒  x24+y212=1.\frac{3x^2}{12} + \frac{y^2}{12} = 1 \;\Rightarrow\; \frac{x^2}{4} + \frac{y^2}{12} = 1.

2. Identify the axes. The larger denominator is under y2y^2 (since 12>412 > 4), so the major axis is along the yy-axis. For an ellipse, aa is always the semi-major axis (the larger value):

a2=12  (a=23),b2=4  (b=2),a^2 = 12 \;(a = 2\sqrt{3}), \qquad b^2 = 4 \;(b = 2),

with bb the semi-minor axis.

3. Apply the latus-rectum formula. For an ellipse, the length of the latus rectum is 2b2a\frac{2b^2}{a}:

L=2b2a=2⋅423=823=43.L = \frac{2b^2}{a} = \frac{2 \cdot 4}{2\sqrt{3}} = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}}. …

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