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NCERT Exemplar · Class 11 Mathematics Conic Sections

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EXERCISE 11.3 51–59 (part 6 of 6)

  1. Choose the correct answer out of the given four options (M.C.Q.) in Exercises $\displaystyle 47$ to 59.

    Exercise 51

    If the focus of a parabola is (0,−3)\displaystyle (0, -3) and its directrix is y=3\displaystyle y=3, then its equation is
    (A)
    x2=−12y\displaystyle x^2=-12 y
    (B)
    x2=12y\displaystyle x^2=12 y
    (C)
    y2=−12x\displaystyle y^2=-12 x
    (D)
    y2=12x\displaystyle y^2=12 x

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    NCERT’s answer
    A
    (A) \(\displaystyle x^2=-12y\) The focus \(\displaystyle (0,-3)\) and the directrix \(\displaystyle y=3\) are symmetric about the origin, so the vertex is \(\displaystyle (0,0)\) and the parabola opens downward. \[a=3 \] \[x^2=-4ay=-12y \]

  2. Exercise 52

    If the parabola y2=4ax\displaystyle y^2=4 a x passes through the point (3,2)\displaystyle (3, 2), then the length of its latus rectum is
    (A)
    23\displaystyle \frac{2}{3}
    (B)
    43\displaystyle \frac{4}{3}
    (C)
    13\displaystyle \frac{1}{3}

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    NCERT’s answer
    B
    (B) \(\displaystyle \tfrac{4}{3}\) Substitute \(\displaystyle (3,2)\) in \(\displaystyle y^2=4ax\): \[4=12a \;\Rightarrow\; a=\tfrac{1}{3} \] \[\text{latus rectum}=4a=\tfrac{4}{3} \]

  3. Exercise 53

    If the vertex of the parabola is the point (−3,0)\displaystyle (-3,0) and the directrix is the line x+5=0\displaystyle x+5=0, then its equation is
    (A)
    y2=8(x+3)\displaystyle y^2=8(x+3)
    (B)
    x2=8(y+3)\displaystyle x^2=8(y+3)
    (C)
    y2=−8(x+3)\displaystyle y^2=-8(x+3)
    (D)
    y2=8(x+5)\displaystyle y^2=8(x+5)

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    NCERT’s answer
    A
    (A) \(\displaystyle y^2=8(x+3)\) The directrix \(\displaystyle x=-5\) lies to the left of the vertex \(\displaystyle (-3,0)\), so the parabola opens right. \[a=-3-(-5)=2 \] \[y^2=4a(x+3)=8(x+3) \]

  4. Exercise 54

    The equation of the ellipse whose focus is (1,−1)\displaystyle (1, -1), the directrix the line x−y−3=0\displaystyle x-y-3=0 and eccentricity 12\displaystyle \frac{1}{2} is
    (A)
    7x2+2xy+7y2−10x+10y+7=0\displaystyle 7 x^2+2 x y+7 y^2-10 x+10 y+7=0
    (B)
    7x2+2xy+7y2+7=0\displaystyle 7 x^2+2 x y+7 y^2+7=0
    (C)
    7x2+2xy+7y2+10x−10y−7=0\displaystyle 7 x^2+2 x y+7 y^2+10 x-10 y-7=0
    (D)
    none

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    NCERT’s answer
    A
    (A) \(\displaystyle 7x^2+2xy+7y^2-10x+10y+7=0\) For \(\displaystyle P(x,y)\): \(\displaystyle PS=e\cdot PM\). \[(x-1)^2+(y+1)^2=\tfrac14\cdot\frac{(x-y-3)^2}{2} \] \[8\left(x^2+y^2-2x+2y+2\right)=x^2+y^2-2xy-6x+6y+9 \] \[7x^2+2xy+7y^2-10x+10y+7=0 \]

  5. Exercise 55

    The length of the latus rectum of the ellipse 3x2+y2=12\displaystyle 3 x^2+y^2=12 is
    (A)
    4\displaystyle 4 (B) 3\displaystyle 3 (C) 8\displaystyle 8 (D) 43\displaystyle \frac{4}{\sqrt{3}}

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    NCERT’s answer
    D
    (D) \(\displaystyle \tfrac{4}{\sqrt3}\) \[3x^2+y^2=12 \;\Rightarrow\; \frac{x^2}{4}+\frac{y^2}{12}=1 \] The larger denominator is under \(\displaystyle y^2\), so the major axis lies along the \(\displaystyle y\)-axis: \(\displaystyle A^2=12,\ B^2=4\). \[\text{latus rectum}=\frac{2B^2}{A}=\frac{8}{2\sqrt3}=\frac{4}{\sqrt3} \]

  6. Exercise 56

    If e\displaystyle e is the eccentricity of the ellipse x2a2+y2b2=1(a<b)\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a<b), then
    (A)
    b2=a2(1−e2)\displaystyle b^2=a^2\left(1-e^2\right)
    (B)
    a2=b2(1−e2)\displaystyle a^2=b^2\left(1-e^2\right)
    (C)
    a2=b2(e2−1)\displaystyle a^2=b^2\left(e^2-1\right)
    (D)
    b2=a2(e2−1)\displaystyle b^2=a^2\left(e^2-1\right)

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    NCERT’s answer
    B
    (B) \(\displaystyle a^2=b^2(1-e^2)\) For \(\displaystyle a<b\) the major axis is along the \(\displaystyle y\)-axis, with semi-major axis \(\displaystyle b\). \[e^2=\frac{b^2-a^2}{b^2} \] \[a^2=b^2\left(1-e^2\right) \]

  7. Exercise 57

    The eccentricity of the hyperbola whose latus rectum is 8\displaystyle 8 and conjugate axis is equal to half of the distance between the foci is
    (A)
    43\displaystyle \frac{4}{3}
    (B)
    43\displaystyle \frac{4}{\sqrt{3}}
    (C)
    23\displaystyle \frac{2}{\sqrt{3}}
    (D)
    none of these

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{2}{\sqrt{3}}\)\[\text{Latus rectum: } \frac{2b^2}{a}=8 \Rightarrow b^2=4a \] \[2b=\tfrac12(2ae) \Rightarrow b=\frac{ae}{2} \] \[b^2=a^2(e^2-1) \Rightarrow \frac{a^2e^2}{4}=a^2(e^2-1) \] \[e^2=4e^2-4 \Rightarrow e^2=\frac{4}{3} \] \[e=\frac{2}{\sqrt{3}} \]

  8. Exercise 58

    The distance between the foci of a hyperbola is 16\displaystyle 16 and its eccentricity is 2\displaystyle \sqrt{2}. Its equation is
    (A)
    x2−y2=32\displaystyle x^2-y^2=32
    (B)
    x24−y29=1\displaystyle \frac{x^2}{4}-\frac{y^2}{9}=1
    (C)
    2x−3y2=7\displaystyle 2 x-3 y^2=7
    (D)
    none of these

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    NCERT’s answer
    A
    (A) \(\displaystyle x^2-y^2=32\)\[2ae=16,\quad e=\sqrt{2} \Rightarrow a=\frac{8}{\sqrt{2}}=4\sqrt{2} \] \[a^2=32 \] \[b^2=a^2(e^2-1)=32(2-1)=32 \] \[\frac{x^2}{32}-\frac{y^2}{32}=1 \Rightarrow x^2-y^2=32 \]

  9. Exercise 59

    Equation of the hyperbola with eccentricty 32\displaystyle \frac{3}{2} and foci at (±2,0)\displaystyle ( \pm 2,0) is
    (A)
    x24−y25=49\displaystyle \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}
    (B)
    x29−y29=49\displaystyle \frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}
    (C)
    x24−y29=1\displaystyle \frac{x^2}{4}-\frac{y^2}{9}=1
    (D)
    none of these

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{x^2}{4}-\dfrac{y^2}{5}=\dfrac{4}{9}\)\[ae=2,\quad e=\frac32 \Rightarrow a=\frac43,\quad a^2=\frac{16}{9} \] \[b^2=a^2(e^2-1)=\frac{16}{9}\cdot\frac54=\frac{20}{9} \] \[\frac{9x^2}{16}-\frac{9y^2}{20}=1 \] \[\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9} \]