The equation of tangent line to $y = 2x^2 + 7$, which is parallel to the line $4x - y + 3 = 0$ is

When two lines are parallel, they have the same slope.

Given line: $4x - y + 3 = 0$

Rearranging to slope-intercept form:

$-y = -4x - 3$

$y = 4x + 3$

The slope of the given line is $4$.

The tangent line must also have slope $4$.

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For the curve $y = 2x^2 + 7$, the derivative gives the slope at any point:

$\frac{dy}{dx} = 4x$

Setting the derivative equal to $4$:

$4x = 4$

$x = 1$

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Substituting $x = 1$ into the original equation:

$y = 2(1)^2 + 7$

$y = 2 + 7$

$y = 9$

The point of tangency is $(1, 9)$.

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Using point-slope form with slope $4$ and point $(1, 9)$:

$y - 9 = 4(x - 1)$

$y - 9 = 4x - 4$

$y = 4x + 5$

$4x - y + 5 = 0$

Therefore, the equation of the tangent line is $4x - y + 5 = 0$.