If the equation of the parabola with vertex $V \left(\frac{3}{2}, 3\right)$ and the directrix $x + 2y = 0$ is $\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$,then $\alpha + \beta + \gamma$ is equal to:

  • A
    $6$
  • B
    $8$
  • C
    $7$
  • D
    $9$

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If the focus of the parabola is $(3, -4)$ and the directrix is $y - 4 = 0,$ then the equation of the parabola is :-

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The line among the following which touches the parabola $y^2=4ax$ is

$A$ line $L: y=mx+3$ meets the $y$-axis at $E(0,3)$ and the arc of the parabola $y^2=16x, 0 \leq y \leq 6$ at the point $F(x_0, y_0)$. The tangent to the parabola at $F(x_0, y_0)$ intersects the $y$-axis at $G(0, y_1)$. The slope $m$ of the line $L$ is chosen such that the area of the triangle $EFG$ has a local maximum.
Match List $I$ with List $II$ and select the correct answer using the code given below the lists:
List $I$ List $II$
$P. \quad m=$ $1. \quad 1/2$
$Q. \quad \text{Maximum area of } \triangle EFG \text{ is}$ $2. \quad 4$
$R. \quad y_0=$ $3. \quad 2$
$S. \quad y_1=$ $4. \quad 1$

Codes: $P \quad Q \quad R \quad S$

$A$ circle is drawn with its centre at the focus of the parabola $y^2 = 2px$ such that it touches the directrix of the parabola. Then a point of intersection of the circle and the parabola is

Let the parabola $y = x^2 + px + q$ passing through the point $(1, -1)$ be such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2 + q^2$ is:

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