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(B) The equation of the parabola is $y^2 = 15x$,so $4a = 15$,which gives $a = \frac{15}{4}$.The point $P$ is $\left(\frac{15}{2}, \frac{15}{\sqrt{2}}\

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(B) The equation of the parabola is $y^2 = 15x$,so $4a = 15$,which gives $a = \frac{15}{4}$.The point $P$ is $\left(\frac{15}{2}, \frac{15}{\sqrt{2}}ight)$. Let $P = (at^2, 2at)$.$at^2 = \frac{15}{2} \implies \frac{15}{4}t^2 = \frac{15}{2} \implies t^2 = 2 \implies t = \sqrt{2}$.The normal at $t$ meets the parabola again at $t_1 = -t - \frac{2}{t} = -\sqrt{2} - \frac{2}{\sqrt{2}} = -2\sqrt{2}$.The vertex is $V(0,0)$. The angle $	heta$ subtended by the chord $PQ$ at the vertex is given by the angle between vectors $\vec{VP}$ and $\vec{VQ}$.$P = (at^2, 2at) = (\frac{15}{2}, \frac{15}{\sqrt{2}})$ and $Q = (at_1^2, 2at_1) = (\frac{15}{4}(8), 2(\frac{15}{4})(-2\sqrt{2})) = (30, -15\sqrt{2})$.The slopes are $m_1 = \frac{2at}{at^2} = \frac{2}{t} = \frac{2}{\sqrt{2}} = \sqrt{2}$ and $m_2 = \frac{2}{t_1} = \frac{2}{-2\sqrt{2}} = -\frac{1}{\sqrt{2}}$.$	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{\sqrt{2} + \frac{1}{\sqrt{2}}}{1 + \sqrt{2}(-\frac{1}{\sqrt{2}})}| = |\frac{3/\sqrt{2}}{0}| = \infty$.Thus,$	heta = 90^\circ$ or $\frac{\pi}{2}$.Substituting $	heta = 90^\circ$ into the expression: $\sin(30^\circ) + \cos(60^\circ) - \sec(120^\circ) = \frac{1}{2} + \frac{1}{2} - (-2) = 1 + 2 = 3$.

If the normal chord drawn at the point $\left(\frac{15}{2}, \frac{15}{\sqrt{2}}\right)$ to the parabola $y^2=15x$ subtends an angle $\theta$ at the vertex of the parabola,then $\sin \frac{\theta}{3}+\cos \frac{2\theta}{3}-\sec \frac{4\theta}{3}=$

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