Let the tangent to the parabola S: y^{2}=2x a | vedclass

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(B) The equation of the parabola is $y^{2}=2x$,so $4a=2 \Rightarrow a=\frac{1}{2}$.The tangent at $P(2,2)$ is given by $yy_{1}=2a(x+x_{1})$.Substituti

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$\frac{25}{2}$

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(B) The equation of the parabola is $y^{2}=2x$,so $4a=2 \Rightarrow a=\frac{1}{2}$.The tangent at $P(2,2)$ is given by $yy_{1}=2a(x+x_{1})$.Substituting $P(2,2)$ and $a=\frac{1}{2}$,we get $2y=1(x+2) \Rightarrow x-2y+2=0$.To find $Q$,set $y=0$ in the tangent equation: $x-2(0)+2=0 \Rightarrow x=-2$. Thus,$Q=(-2,0)$.The slope of the tangent at $P(2,2)$ is $m=\frac{1}{2}$. The slope of the normal at $P$ is $m'=-\frac{1}{m}=-2$.The equation of the normal at $P(2,2)$ is $y-2=-2(x-2) \Rightarrow y=-2x+6$.To find $R$,substitute $y=-2x+6$ into $y^{2}=2x$:$(-2x+6)^{2}=2x$ $\Rightarrow 4x^{2}-24x+36=2x$ $\Rightarrow 4x^{2}-26x+36=0$ $\Rightarrow 2x^{2}-13x+18=0$.$(2x-9)(x-2)=0$. Since $x=2$ is point $P$,the $x$-coordinate of $R$ is $x=\frac{9}{2}$.Then $y=-2(\frac{9}{2})+6=-9+6=-3$. So,$R=(\frac{9}{2}, -3)$.The area of $\Delta PQR$ with vertices $P(2,2)$,$Q(-2,0)$,and $R(\frac{9}{2}, -3)$ is:$	ext{Area} = \frac{1}{2} |x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})|$$= \frac{1}{2} |2(0 - (-3)) + (-2)(-3 - 2) + \frac{9}{2}(2 - 0)|$$= \frac{1}{2} |2(3) + (-2)(-5) + \frac{9}{2}(2)|$$= \frac{1}{2} |6 + 10 + 9| = \frac{1}{2} |25| = \frac{25}{2} \ sq. \ units$.

Let the tangent to the parabola $S: y^{2}=2x$ at the point $P(2,2)$ meet the $x$-axis at $Q$ and the normal at $P$ meet the parabola $S$ at the point $R$. Then the area (in $sq. \ units$) of the triangle $PQR$ is equal to:

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