If $x = t^{2} + 2$ and $y = 2t$ represent the parametric equations of a parabola,find its Cartesian equation.
- A$x^{2} = 4(y - 2)$
- B$(y - 2)^{2} = 4x$
- C$y^{2} = 4(x - 2)$
- D$(x - 2)^{2} = 4y$
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The distance of the point $(6, 4 \sqrt{3})$ from the focus of the parabola $y^2 = 8x$ is:
Find the axis of the parabola $9y^2 - 16x - 12y - 57 = 0$.
Medium
View SolutionThe nearest point on the curve $x^2=2y$ to the point $(0,5)$ is . . . . . . .
For the parabola $y^2+6y-2x+5=0$,match the items in List-$I$ with the suitable item in List-$II$ given below:
List-$I$ List-$II$ $(I)$ Vertex $(A)$ $(-\frac{3}{2}, -3)$ $(II)$ Focus $(B)$ $(\frac{3}{2}, -3)$ $(III)$ Equation of the directrix $(C)$ $2x+5=0$ $(IV)$ Equation of the axis $(D)$ $2x+y+3=0$ $(E)$ $y+3=0$ $(F)$ $(-2, -3)$
| List-$I$ | List-$II$ |
| $(I)$ Vertex | $(A)$ $(-\frac{3}{2}, -3)$ |
| $(II)$ Focus | $(B)$ $(\frac{3}{2}, -3)$ |
| $(III)$ Equation of the directrix | $(C)$ $2x+5=0$ |
| $(IV)$ Equation of the axis | $(D)$ $2x+y+3=0$ |
| $(E)$ $y+3=0$ | |
| $(F)$ $(-2, -3)$ |
$P$ and $Q$ are two distinct points on the parabola,$y^2 = 4x$,with parameters $t$ and $t_1$ respectively. If the normal at $P$ passes through $Q$,then the minimum value of $t_1^2$ is
DifficultJEE MAIN 2016
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