<abstract><p>Let $ q $ be an even prime power and let $ \mathbb{F}_{q} $ be the finite field of $ q $ elements. Let $ f $ be a nonzero polynomial over $ \mathbb{F}_{q^2} $ of the form $ f = a_{1}x_{1}^{m_{1}}+\dots+a_{s}x_{s}^{m_{s}}+y_{1}y_{2}+\dots+y_{n-1}y_{n}+y_{n-2t-1}^{2}+\dots +y_{n-3}^2+y_{n-1}^{2}$ $+b_{t}y_{n-2t}^{2}+ $ $ \dots +b_{1}y_{n-2}^{2}+b_{0}y_{n}^{2} $, where $ a_{i}, b_j\in \mathbb{F}_{q^{2}}^{*}, $ $ m_i\ne 1, $ $ (m_{i}, m_{k}) = 1, $ $ i\ne k, $ $ m_{i}|(q+1), $ $ m_{i}\in \mathbb{Z}^{+}, $ $ 2|n $, $ n &gt; 2 $, $ 0\leq t\leq \frac{n}{2}-2 $, $ {\mathrm{Tr}}_{\mathbb{F}_{q^2}/\mathbb{F}_{2}}(b_{j}) = 1 $ for $ i, k = 1, \dots, s $ and $ j = 0, 1, \dots, t $. For each $ b \in \mathbb{F}_{q^2} $, let $ N_{q^2}(f = b) $ denote the number of $ \mathbb{F}_{q^2} $-rational points on the affine hypersurface $ f = b $. In this paper, we obtain the formula of $ N_{q^2}(f = b) $ by using the Jacobi sums, Gauss sums and the results of quadratic form in finite fields.</p></abstract>