Lines x+y=1 and 3y=x+3 intersect the ellipse | vedclass

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Question

(B) The equation of the ellipse is $x^{2}+9y^{2}=9$,which can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$.First,find the intersection points:$1$

Vedclass · Accepted Answer

$\frac{18}{5}$

Vedclass · Answer

(B) The equation of the ellipse is $x^{2}+9y^{2}=9$,which can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$.First,find the intersection points:$1$. Intersection of $x+y=1$ and $3y=x+3$:Substitute $x=1-y$ into $3y=x+3$: $3y=(1-y)+3$ $\Rightarrow 4y=4$ $\Rightarrow y=1$. Then $x=0$. So,point $P$ is $(0, 1)$.$2$. Intersection of $x+y=1$ and $x^{2}+9y^{2}=9$:Substitute $x=1-y$ into the ellipse equation: $(1-y)^{2}+9y^{2}=9$ $\Rightarrow 1-2y+y^{2}+9y^{2}=9$ $\Rightarrow 10y^{2}-2y-8=0$ $\Rightarrow 5y^{2}-y-4=0$ $\Rightarrow (5y+4)(y-1)=0$. Thus $y=1$ (gives $P(0,1)$) or $y=-4/5$. If $y=-4/5$,$x=1-(-4/5)=9/5$. So,point $R$ is $(9/5, -4/5)$.$3$. Intersection of $3y=x+3$ and $x^{2}+9y^{2}=9$:Substitute $y=(x+3)/3$ into the ellipse equation: $x^{2}+9((x+3)/3)^{2}=9$ $\Rightarrow x^{2}+(x+3)^{2}=9$ $\Rightarrow x^{2}+x^{2}+6x+9=9$ $\Rightarrow 2x^{2}+6x=0$ $\Rightarrow 2x(x+3)=0$. Thus $x=0$ (gives $P(0,1)$) or $x=-3$. If $x=-3$,$y=(-3+3)/3=0$. So,point $Q$ is $(-3, 0)$.The vertices of $	riangle PQR$ are $P(0, 1)$,$Q(-3, 0)$,and $R(9/5, -4/5)$.The area of $	riangle PQR = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$= \frac{1}{2} |0(0 - (-4/5)) + (-3)(-4/5 - 1) + (9/5)(1 - 0)|$$= \frac{1}{2} |0 + (-3)(-9/5) + 9/5|$$= \frac{1}{2} |27/5 + 9/5| = \frac{1}{2} |36/5| = \frac{18}{5}$.

Lines $x+y=1$ and $3y=x+3$ intersect the ellipse $x^{2}+9y^{2}=9$ at the points $P, Q$ and $R$. The area of the $\triangle PQR$ is

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