Maths NCERT Exemplar Solutions Class 12th Chapter Eleven
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New answer posted
2 months agoContributor-Level 10
Equation of tangent of slope m to y = x2 is y = mx
-
Equation of tangent of slope m to y = - (x - 2)2 is y = m (x – 2) +
If both equation represent the same line therefore on comparing (i) and (ii) we get m = 0, 4
therefore equation of tangent is y = 4x – 4
New answer posted
2 months agoContributor-Level 10
Equation of tangent having slope m is
which passes through (1, 3) and we get m1 + m2 = -4 and m1m2 =
Acute angle between the tangents is α = tan-1
New answer posted
2 months agoContributor-Level 10
Normal of plane P : =
Equation of plane P which passes through (2, 2) is 4x – y – 3z – 12 = 0
Now, A (3, 0, 0), B (0, 12 0), C (0, 4)
Now, volume of tetrahedron OABC
(V, P) = (24, 13)
New answer posted
2 months agoContributor-Level 10
Given,
and
are coplanar
Now, normal of plane P, which contains L1 and L2
equation of required plane P : 3x + 13y – 11z + 4 = 0
(0, 4, 5) does not lie on plane P.
New answer posted
2 months agoContributor-Level 10
Foci : S (ae, 0), S' (ae, 0)
Focus of parabola is (ae, 0)
Now, semi latus rectum of parabola = |SS'| = 2ae
B2 = 2a2……… (i)
lies on H
From (i) and (ii)
equation of parabola is y2
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