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NCERT Exemplar · Class 11 Mathematics Conic Sections

59 questions · 59 still being checked

EXERCISE 11.3 41–50 (part 5 of 6)

  1. Fill in the Blank in Exercises from $\displaystyle 41$ to 46.

    Exercise 41

    The equation of the circle having centre at (3,−4)\displaystyle (3,-4) and touching the line 5x+12y−12=0\displaystyle 5 x+12 y-12=0 is ____\displaystyle \_\_\_\_. [Hint: To determine radius find the perpendicular distance from the centre of the circle to the line.]

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    NCERT’s answer
    \(\displaystyle (x-3)^2+(y+4)^2=\left(\frac{45}{13}\right)^2\)
    \(\displaystyle 169x^2+169y^2-1014x+1352y+2200=0\)Radius is the perpendicular distance from the centre to the line.\[r=\frac{|5(3)+12(-4)-12|}{\sqrt{5^2+12^2}}=\frac{45}{13} \] \[(x-3)^2+(y+4)^2=\frac{2025}{169} \] \[169\left(x^2+y^2-6x+8y+25\right)=2025 \] \[169x^2+169y^2-1014x+1352y+2200=0 \]

  2. Exercise 42

    The equation of the circle circumscribing the triangle whose sides are the lines y=x+2,3y=4x,2y=3x\displaystyle y=x+2,3 y=4 x, 2 y=3 x is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle x^2+y^2-46 x+22 y=0\)
    \(\displaystyle x^2+y^2-46x+22y=0\)Vertices of the triangle:\[y=x+2,\ 3y=4x \ \Rightarrow\ (6,8) \] \[y=x+2,\ 2y=3x \ \Rightarrow\ (4,6) \] \[3y=4x,\ 2y=3x \ \Rightarrow\ (0,0) \]The circle passes through the origin.\[x^2+y^2+Dx+Ey=0 \] \[(6,8):\ 100+6D+8E=0 \ \Rightarrow\ 3D+4E=-50 \] \[(4,6):\ 52+4D+6E=0 \ \Rightarrow\ 2D+3E=-26 \] \[E=22,\quad D=-46 \] \[x^2+y^2-46x+22y=0 \]

  3. Exercise 43

    An ellipse is described by using an endless string which is passed over two pins. If the axes are 6\displaystyle 6 cm and 4\displaystyle 4 cm, the length of the string and distance between the pins are ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle 6+2 \sqrt{5}, 2 \sqrt{5}\)
    String length \(\displaystyle 6+2\sqrt5\) cm; distance between pins \(\displaystyle 2\sqrt5\) cm.The pins are the foci. With the pencil at \(\displaystyle P\), the taut string forms triangle \(\displaystyle PF_1F_2\).\[2a=6,\ 2b=4 \ \Rightarrow\ a=3,\ b=2 \] \[c^2=a^2-b^2=9-4=5 \ \Rightarrow\ c=\sqrt5 \] \[F_1F_2=2c=2\sqrt5 \] \[PF_1+PF_2=2a=6 \] \[\text{string}=PF_1+PF_2+F_1F_2=6+2\sqrt5 \]

  4. Exercise 44

    The equation of the ellipse having foci (0,1)\displaystyle (0, 1), (0,−1)\displaystyle (0, -1) and minor axis of length 1\displaystyle 1 is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle \frac{4 x^2}{1}+\frac{4 y^2}{5}=1\)
    \(\displaystyle \dfrac{x^2}{1/4}+\dfrac{y^2}{5/4}=1\), i.e. \(\displaystyle 20x^2+4y^2=5\)Foci lie on the \(\displaystyle y\)-axis, so the major axis is vertical.\[c=1,\quad 2b=1 \ \Rightarrow\ b=\tfrac12 \] \[a^2=b^2+c^2=\tfrac14+1=\tfrac54 \] \[\frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \ \Rightarrow\ \frac{x^2}{1/4}+\frac{y^2}{5/4}=1 \] \[20x^2+4y^2=5 \]

  5. Exercise 45

    The equation of the parabola having focus at (−1,−2)\displaystyle (-1, -2) and the directrix x−2y+3=0\displaystyle x-2 y+3=0 is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle 4 x^2+4 x y+y^2+4 x+32 y+16=0\)
    \(\displaystyle 4x^2+4xy+y^2+4x+32y+16=0\)For a point \(\displaystyle P(x,y)\) on the parabola, distance to focus equals distance to directrix.\[\sqrt{(x+1)^2+(y+2)^2}=\frac{|x-2y+3|}{\sqrt{1^2+2^2}} \] \[5\left[(x+1)^2+(y+2)^2\right]=(x-2y+3)^2 \] \[5x^2+5y^2+10x+20y+25=x^2+4y^2-4xy+6x-12y+9 \] \[4x^2+4xy+y^2+4x+32y+16=0 \]

  6. Exercise 46

    The equation of the hyperbola with vertices at (0,±6)\displaystyle (0, \pm 6) and eccentricity 53\displaystyle \frac{5}{3} is ____\displaystyle \_\_\_\_ and its foci are ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle \frac{y^2}{36}-\frac{x^2}{64}=1\) and \(\displaystyle (0, \pm 10)\).
    \(\displaystyle \dfrac{y^2}{36}-\dfrac{x^2}{64}=1\); foci \(\displaystyle (0,\pm10)\)Vertices are on the \(\displaystyle y\)-axis, so the transverse axis is vertical.\[a=6,\quad e=\frac ca=\frac53 \ \Rightarrow\ c=10 \] \[b^2=c^2-a^2=100-36=64 \] \[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1 \ \Rightarrow\ \frac{y^2}{36}-\frac{x^2}{64}=1 \] \[\text{foci}=(0,\pm c)=(0,\pm10) \]
  7. Choose the correct answer out of the given four options (M.C.Q.) in Exercises $\displaystyle 47$ to 59.

    Exercise 47

    The area of the circle centred at (1,2)\displaystyle (1,2) and passing through (4,6)\displaystyle (4,6) is
    (A)
    5π\displaystyle 5 \pi
    (B)
    10π\displaystyle 10 \pi
    (C)
    25π\displaystyle 25 \pi
    (D)
    none of these

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 25\pi\)The radius is the distance from the centre to the given point.\[r^2=(4-1)^2+(6-2)^2=9+16=25 \] \[\text{Area}=\pi r^2=25\pi \]

  8. Exercise 48

    Equation of a circle which passes through (3,6)\displaystyle (3,6) and touches the axes is
    (A)
    x2+y2+6x+6y+3=0\displaystyle x^2+y^2+6 x+6 y+3=0
    (B)
    x2+y2−6x−6y−9=0\displaystyle x^2+y^2-6 x-6 y-9=0
    (C)
    x2+y2−6x−6y+9=0\displaystyle x^2+y^2-6 x-6 y+9=0
    (D)
    none of these

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    NCERT’s answer
    (C)
    (C) \(\displaystyle x^2+y^2-6x-6y+9=0\)\(\displaystyle (3,6)\) is in the first quadrant, and the circle touches both axes, so the centre is \(\displaystyle (r,r)\) with radius \(\displaystyle r\).NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q48\[(3-r)^2+(6-r)^2=r^2 \] \[r^2-18r+45=0 \ \Rightarrow\ (r-3)(r-15)=0 \] \[r=3 \ \text{(the other root, 15, gives no listed option)} \] \[(x-3)^2+(y-3)^2=9 \] \[x^2+y^2-6x-6y+9=0 \]

  9. Exercise 49

    Equation of the circle with centre on the y\displaystyle y-axis and passing through the origin and the point (2,3)\displaystyle (2,3) is
    (A)
    x2+y2+13y=0\displaystyle x^2+y^2+13 y=0
    (B)
    3x2+3y2+13x+3=0\displaystyle 3 x^2+3 y^2+13 x+3=0
    (C)
    6x2+6y2−13x=0\displaystyle 6 x^2+6 y^2-13 x=0
    (D)
    x2+y2+13x+3=0\displaystyle x^2+y^2+13 x+3=0

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    None of (A)-(D). The circle is \(\displaystyle 3x^2+3y^2-13y=0\). Centre \(\displaystyle (0,k)\); through the origin, so \(\displaystyle r=|k|\). \[x^2+(y-k)^2=k^2 \;\Rightarrow\; x^2+y^2-2ky=0 \] \[(2,3):\quad 4+9-6k=0 \;\Rightarrow\; k=\tfrac{13}{6} \] \[x^2+y^2-\tfrac{13}{3}\,y=0 \;\Rightarrow\; 3x^2+3y^2-13y=0 \] \[\text{(A) at } (2,3):\ 4+9+39=52\neq0 \qquad \text{(C) at } (2,3):\ 24+54-26=52\neq0 \] \[\text{(B), (D) at } (0,0):\ 3\neq0 \] NCERT prints: (C) \(\displaystyle 6x^2+6y^2-13x=0\) -- its centre \(\displaystyle \left(\tfrac{13}{12},0\right)\) is on the \(\displaystyle x\)-axis and it misses \(\displaystyle (2,3)\).

  10. Exercise 50

    The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a\displaystyle 3 a is
    (A)
    x2+y2=9a2\displaystyle x^2+y^2=9 a^2
    (B)
    x2+y2=16a2\displaystyle x^2+y^2=16 a^2
    (C)
    x2+y2=4a2\displaystyle x^2+y^2=4 a^2
    (D)
    x2+y2=a2\displaystyle x^2+y^2=a^2
    [Hint: Centroid of the triangle coincides with the centre of the circle and the radius of the circle is 23\displaystyle \frac{2}{3} of the length of the median]

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    NCERT’s answer
    (C)
    (C) \(\displaystyle x^2+y^2=4a^2\) The centroid of the triangle is the centre \(\displaystyle O\), and it divides each median \(\displaystyle 2:1\). \[r=\tfrac{2}{3}\cdot 3a=2a \] \[x^2+y^2=r^2=4a^2 \]