Position of the point relative to Parabola:
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Consider the parabola : y2 = 4ax. If (x1 , y1) is the given point and y21 -4ax1 = 0, then point lies on the parabola. But when y12 - 4ax1 ≠ 0, we draw ordinate PM meeting the curve in L. Then P will lie outside the parabola if
PM > LM, that is, PM2 - LM2 > 0
Now, PM2 = y12 and LM2 = 4ax1 by virtue of coordinates of L satisfying the equation of parabola. Substituting these values in the equation of parabola, the condition for P to lie outside parabola becomes y12 - 4ax1 > 0.
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Likewise, the condition for P to lie inside parabola is y12 -4ax1 < 0.
Example: The coordinates of a point on parabola y2 = 8x, whose focal distance is 4, are
(A) (1/2, ±2) (B) (1, ±2√2)
(C) (2, ±4) (D) none of these
Solution: Focal distance of the point P (x, y) on y2 = 4ax is (x + a).
∴ 4 = x + 2 => x = 2
y2 = 8 x 2 = 16 => y = ± 4
Hence (C) is the required answer.
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