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Ask QuestionPosted by Nidhi Nidhi 7 years, 9 months ago
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Sia ? 6 years, 6 months ago
Using distance formula, we obtain
SP = {tex}\sqrt { \left( a t ^ { 2 } - a \right) ^ { 2 } + ( 2 a t - 0 ) ^ { 2 } } = a \sqrt { \left( t ^ { 2 } - 1 \right) ^ { 2 } + 4 t ^ { 2 } }{/tex}= a(t2 + 1)
SQ = {tex}\sqrt { \left( \frac { a } { t ^ { 2 } } - a \right) ^ { 2 } + \left( \frac { 2 a } { t } - 0 \right) ^ { 2 } }{/tex}
SQ = {tex}\sqrt { \frac { a ^ { 2 } \left( 1 - t ^ { 2 } \right) ^ { 2 } } { t ^ { 4 } } + \frac { 4 a ^ { 2 } } { t ^ { 2 } } }{/tex}= {tex}\frac { a } { t ^ { 2 } } \sqrt { \left( 1 - t ^ { 2 } \right) ^ { 2 } + 4 t ^ { 2 } } = \frac { a } { t ^ { 2 } } \sqrt { \left( 1 + t ^ { 2 } \right) ^ { 2 } } = \frac { a } { t ^ { 2 } }( 1+ t^2){/tex}
{tex}\therefore \quad \frac { 1 } { S P } + \frac { 1 } { S Q } = \frac { 1 } { a \left( t ^ { 2 } + 1 \right) } + \frac { t ^ { 2 } } { a \left( t ^ { 2 } + 1 \right) }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { S P } + \frac { 1 } { S Q } = \frac { 1 + t ^ { 2 } } { a \left( t ^ { 2 } + 1 \right) } = \frac { 1 } { a }{/tex}, which is independent of t.
Posted by Rakesh Naik 7 years, 9 months ago
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Rakesh Naik 7 years, 9 months ago
Posted by Tushar Bura 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We have to find the least number that is divisible by all numbers between 1 and 10(both inclusive).
The required number is LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
{tex}\therefore{/tex}LCM= 2{tex} \times{/tex}2{tex}\times{/tex} 3{tex} \times{/tex} 2{tex} \times{/tex} 2{tex} \times{/tex} 5{tex} \times{/tex}7 = 2520
Posted by Rakesh Naik 7 years, 9 months ago
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Complicated Life 7 years, 9 months ago
Rakesh Naik 7 years, 9 months ago
Complicated Life 7 years, 9 months ago
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Sia ? 6 years, 6 months ago
The GCD (Greatest Common Divisor) of two numbers is the largest positive integer number that divides both the numbers without leaving any remainder.
Posted by Pranav Chaudhari 7 years, 9 months ago
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Jyotsana Singh 7 years, 9 months ago
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