Parametric point on the parabola whose focus is 0-1 and the directrix is x -2-0 isA- t 2-1-2 t -1B- t2- 2 tC- t 2-1-2 t -1D- t2-1-2 t-1

The correct option is A (t^2 1, 2t+1)Equation of the parabola is x^2+(y 1)^2=(x+2)^2/1⇒ (y 1)^2=4(x+1) ∴ x = t^2 1, y=2t+1 is parametric point. ...

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Standard XII

Mathematics

Double Ordinate

Parametric po...

Question

Parametric point on the parabola whose focus is (0, 1) and the directrix is x + 2 =0 is

(t^{2} + 1, 2 t - 1)

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(t^{2} + 1, 2 t + 1)

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(t^{2}, 2 t)

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(t^{2} - 1, 2 t + 1)

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Solution

The correct option is D $(t^{2} - 1, 2 t + 1)$
Equation of the parabola is
$x^{2} + (y - 1)^{2} = \frac{(x + 2)^{2}}{1} \Rightarrow (y - 1)^{2} = 4 (x + 1)$
$∴ x = t^{2} - 1, y = 2 t + 1$ is parametric point.

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Parametric point on the parabola whose focus is (0, 1) and the directrix is x + 2 =0 is

The correct option is D (t2−1,2t+1)Equation of the parabola is x2+(y−1)2=(x+2)21⇒(y−1)2=4(x+1) ∴x=t2−1,y=2t+1 is parametric point.

The correct option is D $(t^{2} - 1, 2 t + 1)$
Equation of the parabola is
$x^{2} + (y - 1)^{2} = \frac{(x + 2)^{2}}{1} \Rightarrow (y - 1)^{2} = 4 (x + 1)$
$∴ x = t^{2} - 1, y = 2 t + 1$ is parametric point.