\[(a)\;\frac{b^2}{2a}, \tan ^{-1} (b)\quad (b)\;\frac{a^2}{4b},\tan ^{-1}(a) \quad (c)\;\frac{a^2}{b},\tan ^{-1} (2a) \quad (d)\;\frac{2a^2}{b},\tan ^{-1}(a)\]

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$y=ax-bx^2$

for maximum height

$\large\frac{dy}{dx}$$=0$

$a-2bx=0$

$x=\large\frac{a}{2b}$

Therefore $y_{\large max}=a \bigg(\large\frac{a}{2b}\bigg)$$-b\bigg(\large\frac{a}{2b}\bigg)^2=\frac{a^2}{4b}$

angle of projection $\tan {\theta}=> at \;x=0$

$\bigg(\large\frac{dy}{dx}\bigg)_{x=0}=\large\frac{d}{dx}$$[ax-bx^2]\bigg|_{x=0}$

$\quad\qquad\qquad=a-2bx \bigg|_{x=0}$

$\qquad\qquad\quad=a$

Therfore $\tan \theta =a$

$\qquad \theta=\tan ^{-1} (a)$

Hence b is the correct answer.

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