I am having trouble showing the following result.
Let $X$ and $Y$ be two centered, Gaussian random variables with variances $\sigma_X^2$ and $\sigma_Y^2$ (respectively), and correlation coefficient $\rho$. Then $$ \mathbb{E}\{ \exp\left[a X\right] \mid Y=k \} = \exp\left[\frac{a \sigma_X}{2 \sigma_Y} (2 k \rho + a (1-\rho^2) \sigma_X \sigma_Y )\right] $$
I am not aware of a result which presents joint distribution of a log normal and gaussian rv, so i am quite sure how to go by doing this.
edit Based on the hint
We have $$ f_{X\mid Y}(x,k)= \frac{1}{\sqrt{2 \pi } \sqrt{1-\rho ^2} \sigma _x}\exp\left[{-\frac{\left(x-\frac{k \rho \sigma _x}{\sigma _y}\right){}^2}{2 \left(1-\rho ^2\right) \sigma _x^2}}\right]$$ The integral $$I = \frac{1}{\sqrt{2 \pi } \sqrt{1-\rho ^2} \sigma _x}\int \exp\left[a x {-\frac{\left(x-\frac{k \rho \sigma _x}{\sigma _y}\right){}^2}{2 \left(1-\rho ^2\right) \sigma _x^2}}\right] dx $$ which requires the square completion.
hint
You do not need to calculate the joint or conditional density of $e^{aX}$ and $Y$. You just need the conditional density of $X$ on $Y,$ which can be obtained from the bivariate gaussian joint distribution. You can compute your quantity as $$ E(e^{aX}\mid Y=1)=\int e^{ax}f_{X\mid Y}(x,1)dx.$$
The conditional distribution is a gaussian with some mean and variance you can calculate in terms of the other quantities, so this integral can be done by completing the square.