(CIE 0606/2021/m/22/Q4)
The curve $\dfrac{4}{x^{2}}+\dfrac{5}{4 y^{2}}=1 \quad$ and the line $\quad x+2 y=0 \quad$ intersect at two points. Find the exact distance between these points. $\quad[6]$
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(CIE 0606/2021/m/22/Q4)
The curve $\dfrac{4}{x^{2}}+\dfrac{5}{4 y^{2}}=1 \quad$ and the line $\quad x+2 y=0 \quad$ intersect at two points. Find the exact distance between these points. $\quad[6]$
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*********math solution************* $\begin{array}{lll}&x+2 y=0 \Rightarrow x=-2 y \Rightarrow x^{2}=(-2 y)^{2}=4 y^{2} \\&\text { Thus } \begin{array}[t]{lll}&\dfrac{4}{x^{2}}+\dfrac{5}{x^{2}} &=1 \\[5mm]&\dfrac{9}{x^{2}} =1 \\&x^{2} =9 \\&x =\pm 3 \end{array}\\ &\text { when } x=3, y=-\dfrac{x}{2}=-\dfrac{3}{2} .\\& \text { Thus intersection point } P=\left(3,-\dfrac{3}{2}\right) \\&\text { when } x=-3, y=-\dfrac{x}{2}=\dfrac{3}{2} . \\&\text { Thus intersection point } Q=\left(-3,\dfrac{3}{2}\right) \\& PQ=\sqrt{(-3-3)^{2}+\left(\dfrac{3}{2}-\left(-\dfrac{3}{2}\right)\right)^{2}}=\sqrt{36+9}=\sqrt{45}=3 \sqrt{5} .\end{array}$
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