Q.Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum of the parabola $x^2 = -16y$.

Showing the 12 most recent of 17 on this concept.

1. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The coordinates of the focus of the parabola given by the equation $4y-x^{2}+4x-12=0$ are (A) $(2,3)$ (B) $(3,2)$ (C) $(3,3)$ (D) $(3,7)$ (E) $(7,2)$

   ›Reveal solutionSolution

   In standard form $(x-2)^2=4(y-2)$, the focus is one unit above the vertex $(2,2)$, i.e. $(2,3)$.

   Rewrite $4y-x^{2}+4x-12=0$:

   $4y=x^{2}-4x+12=(x-2)^{2}+8\;\Rightarrow\;(x-2)^{2}=4(y-2).$ …

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2. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.The equation of the latus rectum of the parabola $y^{2} + 8x + 4y + 12 = 0$ is (A) $x + 3 = 0$ (B) $y + 3 = 0$ (C) $x + 1 = 0$ (D) $y + 2 = 0$ (E) $x + 2 = 0$

   ›Reveal solutionSolution

   Complete the square to standard form $(y+2)^2=-8(x+1)$; the latus rectum passes through the focus.

   $y^2+8x+4y+12=0\Rightarrow (y+2)^2-4+8x+12=0\Rightarrow (y+2)^2=-8(x+1)$. …

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3. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.The length of the latus rectum of the ellipse $225x^2+125y^2=28125$ is (A) $\frac{50}{3}$ (B) $\frac{25}{3}$ (C) $5$ (D) $\frac{50}{7}$ (E) $\frac{125}{3}$

   ›Reveal solutionSolution

   Dividing by $28125$ gives $\frac{x^2}{125}+\frac{y^2}{225}=1$ with $a^2=225$ (major axis vertical), $b^2=125$. Latus rectum $=\frac{2b^2}{a}=\frac{2\cdot125}{15}=\frac{50}{3}$.

   Divide $225x^2+125y^2=28125$ by $28125$:

   $\frac{x^2}{28125/225}+\frac{y^2}{28125/125}=1\;\Rightarrow\;\frac{x^2}{125}+\frac{y^2}{225}=1.$ …

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4. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.The eccentricity of the hyperbola $25x^2-36y^2-50x-72y-911=0$, is (A) $\frac{\sqrt{61}}{6}$ (B) $\frac{\sqrt{65}}{6}$ (C) $\frac{\sqrt{61}}{4}$ (D) $\frac{\sqrt{59}}{6}$ (E) $\frac{\sqrt{71}}{6}$

   ›Reveal solutionSolution

   Completing the square converts the hyperbola to $\frac{(x-1)^2}{36}-\frac{(y+1)^2}{25}=1$ with $a^2=36,\ b^2=25$. Then $e=\sqrt{1+b^2/a^2}=\sqrt{61/36}=\frac{\sqrt{61}}{6}$.

   Group and complete squares in $25x^2-36y^2-50x-72y-911=0$:

   $25(x^2-2x)-36(y^2+2y)=911.$

   $25[(x-1)^2-1]-36[(y+1)^2-1]=911,$

   $25(x-1)^2-25-36(y+1)^2+36=911,$

   $25(x-1)^2-36(y+1)^2=911+25-36=900.$

   Divide by $900$: …

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5. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The line $x-1=0$ is the directrix of the parabola, $y^{2}-kx+8=0$. Then, the values of $k$ are (A) $8,4$ (B) $8,-4$ (C) $-8,-4$ (D) $8$ (E) $-8,4$

   ›Reveal solutionSolution

   Matching the directrix $x=1$ to the parabola gives $k^2+4k-32=0$, so $k=4$ or $k=-8$.

   Rewrite $y^2-kx+8=0$ as $y^2=kx-8=k\left(x-\dfrac8k\right)$, a parabola with vertex $\left(\dfrac8k,0\right)$ and $4p=k$, so $p=\dfrac k4$. The directrix is $x=\dfrac8k-\dfrac k4$. Setting it to $1$: …

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6. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The length of the latus rectum of $x^{2}=-9y$ is equal to (A) $9$ units (B) $\frac{3}{2}$ units (C) $4$ units (D) $3$ units (E) $\frac{9}{4}$ units

   ›Reveal solutionSolution

   For $x^2=4py$ the latus rectum length is $|4p|$; here $|{-9}|=9$.

   The parabola $x^2=-9y$ is of the form $x^2=4py$ with $4p=-9$. The length of the latus …

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7. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.If the focus and vertex of a parabola are at a distance of 3 units and 6 units, respectively, from the origin on the positive $x$-axis, then the equation of the parabola is (A) $y^{2} = -12(x - 6)$ (B) $y^{2} = -9(x - 6)$ (C) $y^{2} = -16(x - 6)$ (D) $y = 16(x - 6)^{2}$ (E) $y^{2} = 14(x - 6)$

   ›Reveal solutionSolution

   Vertex is farther out than the focus, so the parabola opens toward the origin (leftward).

   Vertex at $(6,0)$, focus at $(3,0)$. The focus lies to the left of the vertex, so the parabola opens in the $-x$ direction. …

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8. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.The focus of the parabola $x^2 - 4x + 8y + 4 = 0$ is (A) $(-2, -2)$ (B) $(1, 1)$ (C) $(2, 1)$ (D) $(2, -2)$ (E) $(1, 2)$

   ›Reveal solutionSolution

   The focus is $(2,-2)$.

   Concept and Intuition

   Complete the square to put the parabola in vertex form $(x-h)^2=-4a(y-k)$; a downward-opening parabola has focus $a$ below the vertex.

   Step-by-Step Solution

   1. $x^2-4x+8y+4=0\Rightarrow (x-2)^2-4+8y+4=0\Rightarrow (x-2)^2=-8y$.
   2. Vertex $(2,0)$; $4a=8\Rightarrow a=2$; opens downward.
   3. Focus $=(2,\,0-2)=(2,-2)$. …

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9. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.Let $y^{2} = 8x$ be the equation of a parabola. Which one of the following is an arbitrary point on the parabola? (A) $(2t, 4t^{2}),\ t \in \mathbb{R}$ (B) $(2t^{2}, 4t^{2}),\ t \in \mathbb{R}$ (C) $(2t^{2}, 2t^{2}),\ t \in \mathbb{R}$ (D) $(2t, 2t^{2}),\ t \in \mathbb{R}$ (E) $(2t^{2}, 4t),\ t \in \mathbb{R}$

   ›Reveal solutionSolution

   Writing $y^2=4ax$ with $4a=8$ gives $a=2$, and the standard parametrization $(at^2,\,2at)=(2t^2,\,4t)$.

   Comparing $y^2=8x$ with $y^2=4ax$ gives $a=2$. A general point of $y^2=4ax$ is $(at^2,\,2at)$. With $a=2$ this is $(2t^2,\,4t)$. …

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10. KEAM 2025Set eng-2025-04284 marksMCQ
    Q.If one end of the latus rectum of the parabola $y^2 = 16x$ is $(4, 8)$, then the coordinates of the other end of the latus rectum, are (A) $(4, -16)$ (B) $(4, 10)$ (C) $(4, -10)$ (D) $(4, 16)$ (E) $(4, -8)$

    ›Reveal solutionSolution

    The two ends of the latus rectum of $y^2=16x$ are $(4,\pm8)$; the other end is $(4,-8)$. …

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11. KEAM 2025Set eng-2025-04294 marksMCQ
    Q.The equation of the directrix of the parabola $(x-1)^2 = 2(y-2)$ is (A) $2y - 3 = 0$ (B) $2y + 3 = 0$ (C) $3y - 2 = 0$ (D) $3y + 2 = 0$ (E) $2x - 1 = 0$

    ›Reveal solutionSolution

    Comparing to $(x-h)^2=4p(y-k)$: $h=1,k=2,4p=2\Rightarrow p=\frac12$; directrix is $y=k-p=\frac32$, i.e. $2y-3=0$. …

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12. KEAM 2024Set eng-2024-06054 marksMCQ
    Q.The length of latus rectum of the parabola $y^2 = x$ is (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{2}$ (C) 4 (D) 1 (E) 2

    ›Reveal solutionSolution

    Latus rectum of $y^2=4ax$ is $4a$; here $4a=1$.

    Write $y^2=x$ as $y^2=4ax$ with $4a=1$, i.e. $a=\tfrac14$. …

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