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Find Equation of the parabola whose vertex is at (0, 0) axis is y axis and passes Through \((\frac{1}{2}, 2)\) also find equation of directrix. |
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Answer» It is given that vertex of the parabola V(0,0)and symmetric about y – axis so its equation is x2 = 4ay or x2 = – 4ay, since the parabola passess through the point \(p(\frac{1}{2}, 2)\) so it lies in the 1st quadrant. ∴ It is an upward parabola x2 = 4ay ∴ It passess though the point \(p(\frac{1}{2}, 2)\) we have \((\frac{1}{2})^2 = 4 \times a\times2\) ⇒ a = \(\frac{1}{32}\) ∴ Required parabola is x2 = 4 ⇒ x2 = \(\frac{1}{8}y\) |
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