Q.For the parabola $x^2 = 6y$, the focus and the equation of directrix are respectively

Showing the 12 most recent of 24 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The co-ordinates of the focus of the parabola $x^2 = -16y$ are

   (a) $(0, -4)$

   (b) $(0, 4)$

   (c) $(4, 0)$

   (d) $(-4, 0)$

   ›Reveal solutionSolution

   A downward-opening parabola $x^2=-4ay$ has its focus at $(0,-a)$; match coefficients to find $a$.

   The standard form of a downward-opening parabola is $x^2 = -4ay$, with focus at $(0,-a)$ and directrix $y=a$.

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 mark
   Q.The coordinates of the focus of the parabola $y^2 = 4ax$ be ______.

   ›Reveal solutionSolution

   The focus of $y^2=4ax$ is $(a,0)$.

   The standard form $y^2=4ax$ represents a parabola opening rightward with vertex at the origin. Comparing with the general standard form, the focus is located on the positive $x$- …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2026Set ANNUAL1 mark
   Q.(Same parabolic-reflector case study as 37(i).) Find the coordinates of the focus F.

   ›Reveal solutionSolution

   From the case study's equation y² = 20x = 4ax, the focus of a rightward-opening parabola is at (a, 0), giving a = 5.

   From the same case study, the parabola's equation was found to be $y^2=20x$, matching the standard form $y^2=4ax$ with:

   $4a = 20 \implies a = 5$ …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2026Set ANNUAL1 mark
   Q.(Same parabolic-reflector case study as 37(i).) What is the length of the latus rectum?

   ›Reveal solutionSolution

   For y² = 4ax, the latus rectum length is always 4a; with a = 5, this gives 20 cm.

   For the parabola $y^2=4ax$ (here $a=5$, from the same case study), the length of the latus rectum is:

   $4a = 4(5) = 20$

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2026Set 1A1 mark
   Q.Find the length of transverse axis of hyperbola $x^2 - 4y^2 = 4$.

   ›Reveal solutionSolution

   $x^2-4y^2=4$ becomes $\tfrac{x^2}{4}-\tfrac{y^2}{1}=1$, so $a=2$ and transverse axis $=2a=4$.

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following is an equation of a parabola?

   (a) $x^2 = 4ay$

   (b) $x = 4ay$

   (c) $4y + x = c$

   (d) $x^2 + 4y^2 = c$

   ›Reveal solutionSolution

   $x^2=4ay$ is the equation of a parabola.

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2025Set ANNUAL1 markMCQ
   Q.For the parabola $x^2 = 6y$, the focus and the equation of directrix are respectively

   (a) $F\left(0, \dfrac{-3}{2}\right), y = \dfrac{3}{2}$

   (b) $F\left(0, \dfrac{3}{2}\right), y = \dfrac{3}{2}$

   (c) $F\left(0, \dfrac{3}{2}\right), y = \dfrac{-3}{2}$

   (d) None of these

   ›Reveal solutionSolution

   Matching $x^2=6y$ to the standard upward parabola $x^2=4ay$ gives $a=3/2$, so focus $(0,3/2)$ and directrix $y=-3/2$.

   The standard form of an upward-opening parabola with vertex at the origin is $x^2 = 4ay$, with focus $(0,a)$ and directrix $y=-a$.

   Comparing $x^2 = 6y$ with $x^2 = 4ay$:

   $4a = 6 \implies a = \frac{3}{2}$ …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2025Set sz1 mark
   Q.The equation of the parabola with focus $(0, -3)$ and directrix $y = 3$ is ........................ .

   ›Reveal solutionSolution

   Vertex at the origin (midpoint of focus $(0,-3)$ and directrix $y=3$), opening downward since the focus is below the directrix, gives $x^2=-12y$.

   The vertex of a parabola is equidistant from the focus and the directrix, and lies on the axis joining them. Focus $(0,-3)$ and directrix $y=3$ are both symmetric about $y=0$, so the vertex is at the origin $(0,0)$, and the axis of the parabola is the y-axis.

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The equation of the directrix of the parabola $x^2=5y$ is

   (a) $4y+5=0$

   (b) $2x+5=0$

   (c) $x+2y=0$

   (d) None of these

   ›Reveal solutionSolution

   Match the given parabola to the standard form $x^2=4ay$ to identify $a$, then use the standard directrix formula $y=-a$.

   The standard upward-opening parabola is $x^2 = 4ay$, with directrix $y = -a$.

   Comparing $x^2 = 5y$ with $x^2 = 4ay$:

   $4a = 5 \implies a = \frac{5}{4}$

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2024Set ANNUAL1 markMCQ
    Q.Co-ordinates of the focus of the parabola $x^2 = -8y$ is —

    (a) $(2,0)$

    (b) $(0,2)$

    (c) $(-2,0)$

    (d) $(0,-2)$

    ›Reveal solutionSolution

    For $x^2=-8y$, the focus is $(0,-2)$.

    The standard downward-opening parabola is $x^2=-4ay$, with vertex at the origin and focus at $(0,-a)$. Comparing $x^2=-8y$ with $x^2=-4ay$ gives $4a=8$, so $a=2$. T …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2024Set ANNUAL1 markMCQ
    Q.The vertex of the parabola $y^2 = -4ax$ is:

    (a) $(-9, 0)$

    (b) $(9, 0)$

    (c) $(0, 0)$

    (d) None of these.

    ›Reveal solutionSolution

    The vertex of $y^2=-4ax$ is the origin $(0,0)$.

    The standard forms of a parabola with vertex at the origin and axis along the x-axis are $y^2=4ax$ (opens right) and $y^2=-4ax$ (opens left, for $a>0$). In both cases the vertex — the point where the para …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2024Set hz1 mark
    Q.Directrix of parabola $y^2 = -4x$ is .......

    ›Reveal solutionSolution

    For $y^2 = -4ax$ (a left-opening parabola), the directrix is the vertical line $x = a$; here $a=1$, so the directrix is $x=1$.

    The given parabola is $y^2 = -4x$. Compare this with the standard form $y^2 = -4ax$, which opens to the left with vertex at the origin:

    $4a = 4 \implies a = 1$

    …

    🔒 Start your 14-day free trial to see the full solution →