$ f_{x,y}(x,y) = c \\ 0\le x \le 25 \\ {x^2 \over 25} \le y $
Find Var$[X|Y=16]$
So here is how I proceeded:-
Var$[X|Y=16] = E[{X^2}|Y=16]-(E[X|Y=16])^2 \\ E[X|Y=16] = \int_{0}^{20}xf_x(X|Y=16) \\ Now, f_x(x|y=16)= \frac{f_{x,y}(x,y)}{f_y(Y=16)}\\ \\f_y(Y=y)=\int_{0}^{5\sqrt[]{y}}\, c\, \ dx\\ \\\Rightarrow f_y(Y=16)=20c\\ \\\therefore f_x(x|y=16)= \frac{f_{x,y}(x,y)}{f_y(Y=16)} = \frac{c}{20c} = {1\over20}\\ E[X|Y=16] = \int_{0}^{20}xf_x(X|Y=16) = \int_{0}^{20}x\frac{1}{20}\,\ dx = 10 \\ Similarly,\\ E[X^2|Y=16] = \int_{0}^{20}x^2f_x(X|Y=16) = \int_{0}^{20}x^2\frac{1}{20}\,\ dx= {400\over3}\\ \therefore , Var[X|Y=16] = E[{X^2}|Y=16]-(E[X|Y=16])^2= {400\over3} -100 ={100\over3} $
Is my answer correct ? I am not quite so sure about limits on the integration. Any kind of help shall be appreciated
Note that $f_{X,Y}(x,16)$ is constant on $[0,20]$ and takes value $0$ for $x\notin[0,20]$.
This information tells us that $X$ under condition $Y=16$ has uniform distribution over $[0,20]$.
So it has the same distribution as $10+10U$ where $U$ has uniform distribution over $[-1,1]$.
Observe that here $\mathbb EU=0$ by symmetry so that $\text{Var}(U)=\mathbb EU^2=\frac12\int^{1}_{-1}u^2du=\int^{1}_{-1}u^2du=\frac13$.
Then we find: $$\text{Var}(X\mid Y=16)=\text{Var}(10+10U)=100\text{Var}(U)=100\mathbb EU^2=\frac{100}3$$
This agrees with your answer and illustrates that things can be done more easily.