If the tangent at a point P on the parabola y | vedclass

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(D) For the parabola $y^2=3x$,the slope of the tangent at $P(x_1, y_1)$ is given by $2y \frac{dy}{dx} = 3$,so $\frac{dy}{dx} = \frac{3}{2y}$.Since the

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$3\sqrt{5}$

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(D) For the parabola $y^2=3x$,the slope of the tangent at $P(x_1, y_1)$ is given by $2y \frac{dy}{dx} = 3$,so $\frac{dy}{dx} = \frac{3}{2y}$.Since the tangent is parallel to $x+2y=1$ (slope $= -1/2$),we have $\frac{3}{2y_1} = -1/2$,which gives $y_1 = -3$. Substituting into $y^2=3x$,we get $x_1 = 3$. Thus,$P = (3, -3)$.For the ellipse $\frac{x^2}{4} + y^2 = 1$,the slope of the tangent at $(x, y)$ is $\frac{dy}{dx} = -\frac{x}{4y}$.The tangents at $Q$ and $R$ are perpendicular to $x-y=2$ (slope $= 1$),so their slope is $-1$. Thus,$-\frac{x}{4y} = -1$,which implies $x = 4y$.Substituting $x=4y$ into the ellipse equation: $\frac{(4y)^2}{4} + y^2 = 1 \Rightarrow 4y^2 + y^2 = 1 \Rightarrow 5y^2 = 1 \Rightarrow y = \pm \frac{1}{\sqrt{5}}$.Then $x = \pm \frac{4}{\sqrt{5}}$. So $Q = (\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}})$ and $R = (-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}})$.The area of $	riangle PQR = \frac{1}{2} |x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)|$.Area $= \frac{1}{2} |3(\frac{1}{\sqrt{5}} - (-\frac{1}{\sqrt{5}})) + \frac{4}{\sqrt{5}}(-\frac{1}{\sqrt{5}} - (-3)) + (-\frac{4}{\sqrt{5}})(-3 - \frac{1}{\sqrt{5}})|$.Area $= \frac{1}{2} |3(\frac{2}{\sqrt{5}}) + \frac{4}{\sqrt{5}}(3 - \frac{1}{\sqrt{5}}) - \frac{4}{\sqrt{5}}(-3 - \frac{1}{\sqrt{5}})| = \frac{1}{2} |\frac{6}{\sqrt{5}} + \frac{12}{\sqrt{5}} - \frac{4}{5} + \frac{12}{\sqrt{5}} + \frac{4}{5}| = \frac{1}{2} |\frac{30}{\sqrt{5}}| = 3\sqrt{5}$.

If the tangent at a point $P$ on the parabola $y^2=3x$ is parallel to the line $x+2y=1$ and the tangents at the points $Q$ and $R$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ are perpendicular to the line $x-y=2$,then the area of the triangle $PQR$ is:

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