If A at 2- 2 at - B a - t 2- -2 a - t and C a - 0- then 2 a is equal toA- A-M- of C A and C BB- H-M- of C A and C BC- G-M- of C A and C BD- None of these

The correct option is B H.M. of CA and CB CA = √((at^2 a)^2 + (2at)^2) = a√((t^2 1)^2 + 4t^2) = a√((t^2 + 1 + 2t^2)) = a(1 + t^2) CB = √((a/t^2 a)^ ...

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Byju's Answer

Standard XI

Mathematics

Cube Root of a Complex Number

If A at 2, 2 ...

Question

If $A (a t^{2}, 2 a t), B (\frac{a}{t^{2}}, \frac{- 2 a}{t})$ and C(a, 0), then 2a is equal to

A.M. of CA and CB

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G.M. of CA and CB

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H.M. of CA and CB

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None of these

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Solution

The correct option is C
H.M. of CA and CB

CA = $\sqrt{(a t^{2} - a)^{2} + (2 a t)^{2}} = a \sqrt{(t^{2} - 1)^{2} + 4 t^{2}}$

= $a \sqrt{(t^{2} + 1 + 2 t^{2})} = a (1 + t^{2})$

CB = $\sqrt{{(\frac{a}{t^{2}} - a)}^{2} + {(\frac{- 2 a}{t})}^{2}} = a (1 + \frac{1}{t^{2}})$

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(JEE Advanced, 2013)

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If A(at2,2at),B(at2,−2at) and C(a, 0), then 2a is equal to

The correct option is C H.M. of CA and CB CA = √(at2−a)2+(2at)2=a√(t2−1)2+4t2 = a√(t2+1+2t2)=a(1+t2) CB = √(at2−a)2+(−2at)2=a(1+1t2)

If $A (a t^{2}, 2 a t), B (\frac{a}{t^{2}}, \frac{- 2 a}{t})$ and C(a, 0), then 2a is equal to

The correct option is C
H.M. of CA and CB

CA = $\sqrt{(a t^{2} - a)^{2} + (2 a t)^{2}} = a \sqrt{(t^{2} - 1)^{2} + 4 t^{2}}$

= $a \sqrt{(t^{2} + 1 + 2 t^{2})} = a (1 + t^{2})$

CB = $\sqrt{{(\frac{a}{t^{2}} - a)}^{2} + {(\frac{- 2 a}{t})}^{2}} = a (1 + \frac{1}{t^{2}})$