If G is the geometric mean of x and y, then tG2−x21+G2−y21=G21

G2=xy∴LHS= G2−x21+G2−y21=xy−x21+xy−y21=x(y−x)1−y(y−x)1=y−x1[x1−y1 ]=y−x1[xyy−x ]=xy1=G21=RHSHence, proved

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If G is the geometric mean of x and y, then t{1}{{{G^2} - {x^2

Question

If G is the geometric mean of x and y, then
$t \frac{1}{G ^{2} - x ^{2}} + \frac{1}{G ^{2} - y ^{2}} = \frac{1}{G ^{2}}$

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Harper Discussed

If G is the geometric mean of x and y, then

t \frac{1}{G ^{2} - x ^{2}} + \frac{1}{G ^{2} - y ^{2}} = \frac{1}{G ^{2}}

5 mins ago

Discuss this question LIVE

5 mins ago

G^{2} = x y

∴

LHS=

\frac{1}{G ^{2} - x ^{2}} + \frac{1}{G ^{2} - y ^{2}}

= \frac{1}{x y - x ^{2}} + \frac{1}{x y - y ^{2}}

= \frac{1}{x ( y - x )} - \frac{1}{y ( y - x )}

= \frac{1}{y - x} [\frac{1}{x} - \frac{1}{y}]

= \frac{1}{y - x} [\frac{y - x}{x y}]

= \frac{1}{x y}

= \frac{1}{G ^{2}} =

RHS

Hence, proved

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Question Text	If G is the geometric mean of x and y, then $t \frac{1}{G ^{2} - x ^{2}} + \frac{1}{G ^{2} - y ^{2}} = \frac{1}{G ^{2}}$
Answer Type	Text solution:1
Upvotes	150