LE the common tangents to the curves and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and respectively denote the eccentricity and the length of the latus rectum of this ellipse, then is equal to
LE the common tangents to the curves and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and respectively denote the eccentricity and the length of the latus rectum of this ellipse, then is equal to
Let y = mx + c is the common tangent
so equation of common tangents will be
which intersects at Q (3, 0)
Major axis and minor axis of ellipse are 12 and 6. So eccentricity
and length of latus rectum
Hence
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centre
radius
centre
OA =
r2 =
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This is a True or False Type Questions as classified in NCERT Exemplar
30.
30. For inequality y+8 ≤ 2x, the equation of the line is y+8=2x. We consider the table below to plot of y+8=2x.
The line devides the xy-plane into half planer I and II. We select a point (0,0) and check the correctness of the inequality.
i.e., 0+8 ≤ 2 × 0
0 ≤ 0 which is true.
So, the solu
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Maths NCERT Exemplar Solutions Class 12th Chapter Six 2025
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