Q.A parabola has its vertex at the origin; its axis lies along the $x$-axis and it passes through the point P(2, 3). The equation of the parabola is

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 $y=a$.

   …

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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 $x$- …

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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:

   $4a = 20 \implies a = 5$ …

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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:

   $4a = 4(5) = 20$

   …

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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 $=2a=4$.

   …

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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.

   …

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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 $x^2 = 4ay$:

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

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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.

   …

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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 $x^2 = 4ay$:

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

   …

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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 …

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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 …

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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:

    $4a = 4 \implies a = 1$

    …

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