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(D) The equation of the parabola is $y^2 = 9x$,so $4a = 9$,which gives $a = \frac{9}{4}$.Let the slope of the tangent be $m$. The equation of the tang

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$\frac{\pi}{4} < 	heta < \frac{\pi}{3}$

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(D) The equation of the parabola is $y^2 = 9x$,so $4a = 9$,which gives $a = \frac{9}{4}$.Let the slope of the tangent be $m$. The equation of the tangent is $y = mx + \frac{a}{m} = mx + \frac{9}{4m}$.Since the tangent passes through $(1, 5)$,we have $5 = m(1) + \frac{9}{4m}$.Multiplying by $4m$,we get $20m = 4m^2 + 9$,or $4m^2 - 20m + 9 = 0$.Let the roots be $m_1$ and $m_2$. Then $m_1 + m_2 = 5$ and $m_1 m_2 = \frac{9}{4}$.The angle $	heta$ between the tangents is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.$|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2} = \sqrt{25 - 4(\frac{9}{4})} = \sqrt{25 - 9} = \sqrt{16} = 4$.Thus,$	an 	heta = |\frac{4}{1 + 9/4}| = |\frac{4}{13/4}| = \frac{16}{13}$.Since $	an \frac{\pi}{4} = 1$ and $	an \frac{\pi}{3} = \sqrt{3} \approx 1.732$,and $1 < \frac{16}{13} < 1.732$,we have $\frac{\pi}{4} < 	heta < \frac{\pi}{3}$.

If $\theta$ is the acute angle between the tangents drawn from the point $(1,5)$ to the parabola $y^2=9x$,then:

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