If $a$ is sufficiently large as compared with $b,$ and $\sqrt \frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the value of $k?$ 
If  $a$ is sufficiently large as compared with $b,$ and $\sqrt
 \frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the
value of $k?$

Query
Can we find it using particular value?,like putting $a=4,b=2,$ then we get

$k=4\sqrt2+4\sqrt \frac{2}{3}-8$

This problem is given in  Multiple Choice Question exercise,the possible answers given are
A.$\frac{2}{3}$
B.$\frac{3}{4}$
C.$\frac{4}{5}$
D.$\frac{5}{6}$
None of the answer mathches with mine....
Please give some hint to determine value of $k?$

Use the Negative Binomial Theorem.
$(1-a/b)^{-1/2}+(1+a/b)^{-1/2}=2+2\left(\frac{a}b\right)^2\left[\binom{-1/2}2+\binom{-1/2}4\left(\frac{a}b\right)^2+\binom{-1/2}6\left(\frac{a}b\right)^4+...\right]$
Since we want $k$ (constant) to be independent of $a/b$, we reject the higher order terms in the bracket on the ground that $a>>b$, giving $k=2\binom{-1/2}2=(-1/2)\times(-1/2-1)=3/4$.

Note that with the obtained $k$, $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}\approx2+k(a/b)^2$ i.e. $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ is not actually equal to $2+k(a/b)^2$ in general. As you have seen, for equality $k$ will have to be taken as a function of $a,b$ and not constant. Thus plugging in specific values of $a,b$ to find $k$ (which gives the exact value of $k$) will not yield the desired answer.
If you wanted to take a risk, you could check which option for $k$ yields the closest answer to the value of $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ for the selected $a,b$. But remember that you still might not get $k=3/4$ since it is not necessary that $3/4$ gives the closest approximation of $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ for all $a,b$.
