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Exercise 7.1 · Q15

Q.Find coordinate of the point on the parabola 2y2=7x2y^2=7x whose parameter is −2-2. Also find focal distance.

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2y2=7x⇒y2=72x⇒4a=72⇒a=782y^2=7x \Rightarrow y^2=\dfrac{7}{2}x \Rightarrow 4a=\dfrac{7}{2}\Rightarrow a=\dfrac{7}{8}.

At t=−2t=-2: x=at2=78×4=72x=at^2=\dfrac{7}{8}\times4=\dfrac{7}{2}, y=2at=2×78×(−2)=−72y=2at=2\times\dfrac{7}{8}\times(-2)=-\dfrac{7}{2}.

Point: (72,−72)\left(\dfrac{7}{2},-\dfrac{7}{2}\right). …

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