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Exercise 7.1 · Q11

Q.For the parabola 3y2=16x3y^2 = 16x, find the parameter of the point (3,−4)(3,-4).

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3y2=16x⇒y2=163x3y^2=16x \Rightarrow y^2=\dfrac{16}{3}x, so 4a=163⇒a=434a=\dfrac{16}{3}\Rightarrow a=\dfrac{4}{3}. …

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