Find the equation of the tangent at the verte | vedclass

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(A) Given the equation of the parabola: $4y^2 + 6x = 8y + 7$. Rearrange the terms to complete the square for $y$: $4(y^2 - 2y) = -6x + 7$. Add $4(1)^2

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$x = 11/6$

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(A) Given the equation of the parabola: $4y^2 + 6x = 8y + 7$. Rearrange the terms to complete the square for $y$: $4(y^2 - 2y) = -6x + 7$. Add $4(1)^2 = 4$ to both sides: $4(y^2 - 2y + 1) = -6x + 7 + 4$. $4(y - 1)^2 = -6x + 11$. Divide by $4$: $(y - 1)^2 = -\frac{6}{4}(x - \frac{11}{6})$. $(y - 1)^2 = -\frac{3}{2}(x - \frac{11}{6})$. This is in the form $(y - k)^2 = 4a(x - h)$,where the vertex $(h, k)$ is $(\frac{11}{6}, 1)$. The tangent at the vertex of a parabola is a line perpendicular to the axis of the parabola passing through the vertex. Since the parabola is of the form $(y - k)^2 = 4a(x - h)$,its axis is horizontal $(y = k)$. Therefore,the tangent at the vertex is a vertical line passing through the vertex $(\frac{11}{6}, 1)$. The equation of this vertical line is $x = h$,which is $x = \frac{11}{6}$.

Find the equation of the tangent at the vertex of the parabola $4y^2 + 6x = 8y + 7$.

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