SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Conic Sections

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EXERCISE 10.2 1–12 (part 3 of 8)

  1. In each of the following Exercises $\displaystyle 1$ to $\displaystyle 6$, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

    Exercise 1

    y2=12x\displaystyle y^{2}=12 x

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    NCERT’s answer
    $\displaystyle \mathrm{F}(3,0)$, axis - $\displaystyle x$ - axis, directrix $\displaystyle x=-3$, length of the Latus rectum = $\displaystyle 12$
    Standard parabola \(\displaystyle y^{2}=4ax\). For \(\displaystyle y^{2}=4ax\ (a>0)\) the parabola opens to the right: focus \(\displaystyle (a,0)\), axis the \(\displaystyle x\)-axis, directrix \(\displaystyle x=-a\), latus rectum of length \(\displaystyle 4a\).Comparing \(\displaystyle y^{2}=12x\) with \(\displaystyle y^{2}=4ax\): \[4a=12\ \Longrightarrow\ a=3.\]Focus \(\displaystyle (3,0)\); axis: the \(\displaystyle x\)-axis \(\displaystyle (y=0)\); directrix \(\displaystyle x=-3\); length of latus rectum \(\displaystyle =12\).

  2. Exercise 2

    x2=6y\displaystyle x^{2}=6 y

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    NCERT’s answer
    $\displaystyle \mathrm{F}\left(0, \frac{3}{2}\right)$, axis $\displaystyle -y$-axis, directrix $\displaystyle y=-\frac{3}{2}$, length of the Latus rectum = $\displaystyle 6$
    Standard parabola \(\displaystyle x^{2}=4ay\). For \(\displaystyle x^{2}=4ay\ (a>0)\) the parabola opens upward: focus \(\displaystyle (0,a)\), axis the \(\displaystyle y\)-axis, directrix \(\displaystyle y=-a\), latus rectum of length \(\displaystyle 4a\).Comparing \(\displaystyle x^{2}=6y\) with \(\displaystyle x^{2}=4ay\): \[4a=6\ \Longrightarrow\ a=\frac32.\]Focus \(\displaystyle \left(0,\dfrac32\right)\); axis: the \(\displaystyle y\)-axis \(\displaystyle (x=0)\); directrix \(\displaystyle y=-\dfrac32\); length of latus rectum \(\displaystyle =6\).

  3. Exercise 3

    y2=−8x\displaystyle y^{2}=-8 x

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    NCERT’s answer
    $\displaystyle \mathrm{F}(-2,0)$, axis - $\displaystyle x$ - axis, directrix $\displaystyle x=2$, length of the Latus rectum = $\displaystyle 8$
    Standard parabola \(\displaystyle y^{2}=-4ax\). For \(\displaystyle y^{2}=-4ax\ (a>0)\) the parabola opens to the left: focus \(\displaystyle (-a,0)\), axis the \(\displaystyle x\)-axis, directrix \(\displaystyle x=a\), latus rectum of length \(\displaystyle 4a\).Comparing \(\displaystyle y^{2}=-8x\) with \(\displaystyle y^{2}=-4ax\): \[4a=8\ \Longrightarrow\ a=2.\]Focus \(\displaystyle (-2,0)\); axis: the \(\displaystyle x\)-axis \(\displaystyle (y=0)\); directrix \(\displaystyle x=2\); length of latus rectum \(\displaystyle =8\).

  4. Exercise 4

    x2=−16y\displaystyle x^{2}=-16 y

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    NCERT’s answer
    $\displaystyle \mathrm{F}(0,-4)$, axis $\displaystyle -y$-axis, directrix $\displaystyle y=4$, length of the Latus rectum = $\displaystyle 16$
    Standard form of a parabola. Compare \(\displaystyle x^{2}=-16y\) with the standard form \(\displaystyle x^{2}=-4ay\), which is the parabola with vertex at the origin opening downwards; for it the focus is \(\displaystyle (0,-a)\), the axis is the \(\displaystyle y\)-axis, the directrix is \(\displaystyle y=a\) and the latus rectum has length \(\displaystyle 4a\).Here \[4a=16 \quad\Rightarrow\quad a=4 . \]Hence the focus is \(\displaystyle (0,-4)\), the directrix is \(\displaystyle y=4\) and the latus rectum is \(\displaystyle 4a=16\).Focus \(\displaystyle (0,-4)\); axis: the \(\displaystyle y\)-axis; directrix \(\displaystyle y=4\); length of latus rectum \(\displaystyle =16\).

  5. Exercise 5

    y2=10x\displaystyle y^{2}=10 x

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    NCERT’s answer
    $\displaystyle \mathrm{F}\left(\frac{5}{2}, 0\right)$ axis - $\displaystyle x$ - axis, directrix $\displaystyle x=-\frac{5}{2}$, length of the Latus rectum = $\displaystyle 10$
    Standard form of a parabola. Compare \(\displaystyle y^{2}=10x\) with the standard form \(\displaystyle y^{2}=4ax\), the parabola with vertex at the origin opening to the right; for it the focus is \(\displaystyle (a,0)\), the axis is the \(\displaystyle x\)-axis, the directrix is \(\displaystyle x=-a\) and the latus rectum has length \(\displaystyle 4a\).Here \[4a=10 \quad\Rightarrow\quad a=\frac{5}{2}. \]Hence the focus is \(\displaystyle \left(\dfrac{5}{2},\,0\right)\), the directrix is \(\displaystyle x=-\dfrac{5}{2}\) and the latus rectum is \(\displaystyle 4a=10\).Focus \(\displaystyle \left(\tfrac{5}{2},0\right)\); axis: the \(\displaystyle x\)-axis; directrix \(\displaystyle x=-\tfrac{5}{2}\); length of latus rectum \(\displaystyle =10\).

  6. Exercise 6

    x2=−9y\displaystyle x^{2}=-9 y

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    NCERT’s answer
    $\displaystyle \mathrm{F}\left(0, \frac{-9}{4}\right)$, axis $\displaystyle -y$-axis, directrix $\displaystyle y=\frac{9}{4}$, length of the Latus rectum $\displaystyle =9$
    Standard form of a parabola. Compare \(\displaystyle x^{2}=-9y\) with the standard form \(\displaystyle x^{2}=-4ay\), the parabola with vertex at the origin opening downwards; for it the focus is \(\displaystyle (0,-a)\), the axis is the \(\displaystyle y\)-axis, the directrix is \(\displaystyle y=a\) and the latus rectum has length \(\displaystyle 4a\).Here \[4a=9 \quad\Rightarrow\quad a=\frac{9}{4}. \]Hence the focus is \(\displaystyle \left(0,\,-\dfrac{9}{4}\right)\), the directrix is \(\displaystyle y=\dfrac{9}{4}\) and the latus rectum is \(\displaystyle 4a=9\).Focus \(\displaystyle \left(0,-\tfrac{9}{4}\right)\); axis: the \(\displaystyle y\)-axis; directrix \(\displaystyle y=\tfrac{9}{4}\); length of latus rectum \(\displaystyle =9\).
  7. In each of the Exercises $\displaystyle 7$ to $\displaystyle 12$, find the equation of the parabola that satisfies the given conditions:

    Exercise 7

    Focus (6,0)\displaystyle (6,0); directrix x=−6\displaystyle x=-6

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    NCERT’s answer
    $\displaystyle y^{2}=24 x$
    Locate the vertex first. The vertex of a parabola is the midpoint of the perpendicular from the focus to the directrix. The focus \(\displaystyle (6,0)\) lies on the \(\displaystyle x\)-axis and the directrix \(\displaystyle x=-6\) is perpendicular to it, so the foot of that perpendicular is \(\displaystyle (-6,0)\) and the vertex is \[\left(\frac{6+(-6)}{2},\;\frac{0+0}{2}\right)=(0,0). \]The axis is the \(\displaystyle x\)-axis and the focus lies to the right of the vertex, so the parabola is of the form \(\displaystyle y^{2}=4ax\) with \(\displaystyle a=6\) (the distance from vertex to focus).\[y^{2}=4(6)x . \]Check: for \(\displaystyle y^{2}=24x\) the directrix is \(\displaystyle x=-a=-6\), as required.\(\displaystyle y^{2}=24x\)

  8. Exercise 8

    Focus (0,−3)\displaystyle (0,-3); directrix y=3\displaystyle y=3

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    NCERT’s answer
    $\displaystyle x^{2}=-12 y$
    Locate the vertex first. The vertex is the midpoint of the perpendicular from the focus to the directrix. The focus \(\displaystyle (0,-3)\) lies on the \(\displaystyle y\)-axis and the directrix \(\displaystyle y=3\) meets that axis at \(\displaystyle (0,3)\), so the vertex is \[\left(\frac{0+0}{2},\;\frac{-3+3}{2}\right)=(0,0). \]The axis is the \(\displaystyle y\)-axis and the focus lies below the vertex, so the parabola is of the form \(\displaystyle x^{2}=-4ay\) with \(\displaystyle a=3\).\[x^{2}=-4(3)y . \]Check: for \(\displaystyle x^{2}=-12y\) the directrix is \(\displaystyle y=a=3\), as required.\(\displaystyle x^{2}=-12y\)

  9. Exercise 9

    Vertex (0,0)\displaystyle (0,0); focus (3,0)\displaystyle (3,0)

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    NCERT’s answer
    $\displaystyle y^{2}=12 x$
    Standard form from vertex and focus. The vertex is at the origin and the focus \(\displaystyle (3,0)\) lies on the positive \(\displaystyle x\)-axis, so the axis of the parabola is the \(\displaystyle x\)-axis and it opens to the right. Such a parabola has the equation \[y^{2}=4ax , \] where \(\displaystyle a\) is the distance from the vertex to the focus.Here \(\displaystyle a=3\), so \[y^{2}=4(3)x . \]\(\displaystyle y^{2}=12x\)

  10. Exercise 10

    Vertex (0,0)\displaystyle (0,0); focus (−2,0)\displaystyle (-2,0)

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    NCERT’s answer
    $\displaystyle y^{2}=-8 x$
    Standard form from vertex and focus. The vertex is at the origin and the focus \(\displaystyle (-2,0)\) lies on the negative \(\displaystyle x\)-axis, so the axis is the \(\displaystyle x\)-axis and the parabola opens to the left. Such a parabola has the equation \[y^{2}=-4ax , \] where \(\displaystyle a\) is the distance from the vertex to the focus.Here \(\displaystyle a=2\), so \[y^{2}=-4(2)x . \]\(\displaystyle y^{2}=-8x\)

  11. Exercise 11

    Vertex (0,0)\displaystyle (0,0) passing through (2,3)\displaystyle (2,3) and axis is along x\displaystyle x-axis.

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    NCERT’s answer
    $\displaystyle 2 y^{2}=9 x$
    Choose the standard form, then fit the point. The vertex is \(\displaystyle (0,0)\) and the axis is along the \(\displaystyle x\)-axis, so the equation is \(\displaystyle y^{2}=4ax\) (with \(\displaystyle 4a\) of either sign). The parabola passes through \(\displaystyle (2,3)\), where both coordinates are positive, so it opens to the right.Substituting \(\displaystyle x=2,\;y=3\): \[3^{2}=4a(2)\quad\Rightarrow\quad 9=8a\quad\Rightarrow\quad a=\frac{9}{8}. \]Therefore \[y^{2}=4\left(\frac{9}{8}\right)x=\frac{9}{2}x . \]Check: at \(\displaystyle (2,3)\), LHS \(\displaystyle =9\) and RHS \(\displaystyle =\tfrac{9}{2}\times 2=9\), so the point lies on the curve.\(\displaystyle 2y^{2}=9x\)

  12. Exercise 12

    Vertex (0,0)\displaystyle (0,0), passing through (5,2)\displaystyle (5,2) and symmetric with respect to y\displaystyle y-axis.

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    NCERT’s answer
    $\displaystyle 2 x^{2}=25 y$
    Choose the standard form, then fit the point. The vertex is \(\displaystyle (0,0)\) and the curve is symmetric about the \(\displaystyle y\)-axis, so the \(\displaystyle y\)-axis is its axis and the equation is \(\displaystyle x^{2}=4ay\) (with \(\displaystyle 4a\) of either sign). The parabola passes through \(\displaystyle (5,2)\), whose \(\displaystyle y\)-coordinate is positive, so it opens upwards.Substituting \(\displaystyle x=5,\;y=2\): \[5^{2}=4a(2)\quad\Rightarrow\quad 25=8a\quad\Rightarrow\quad a=\frac{25}{8}. \]Therefore \[x^{2}=4\left(\frac{25}{8}\right)y=\frac{25}{2}y . \]Check: at \(\displaystyle (5,2)\), LHS \(\displaystyle =25\) and RHS \(\displaystyle =\tfrac{25}{2}\times 2=25\), so the point lies on the curve.\(\displaystyle 2x^{2}=25y\)