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NCERT Exemplar · Class 11 Mathematics Conic Sections

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EXERCISE 11.3 21–30 (part 3 of 6)

  1. Exercise 21

    Find the eccentricity of the hyperbola 9y2−4x2=36\displaystyle 9 y^2-4 x^2=36.

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    NCERT’s answer
    \(\displaystyle \frac{\sqrt{13}}{2}\)
    The \(\displaystyle y^2\) term is positive, so the transverse axis is the \(\displaystyle y\)-axis. \[9y^2-4x^2=36 \Rightarrow \frac{y^2}{4}-\frac{x^2}{9}=1 \] \[a^2=4,\quad b^2=9 \] \[e^2=1+\frac{b^2}{a^2}=1+\frac94=\frac{13}{4} \] Answer: \(\displaystyle e=\dfrac{\sqrt{13}}{2}\)

  2. Exercise 22

    Find the equation of the hyperbola with eccentricity 32\displaystyle \frac{3}{2} and foci at (±2,0)\displaystyle ( \pm 2,0).

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    NCERT’s answer
    \(\displaystyle \frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}\).
    The foci lie on the \(\displaystyle x\)-axis, so the centre is the origin and the transverse axis is the \(\displaystyle x\)-axis. \[ae=2,\quad e=\tfrac32 \Rightarrow a=\tfrac43,\quad a^2=\tfrac{16}{9} \] \[b^2=a^2(e^2-1)=\tfrac{16}{9}\left(\tfrac94-1\right)=\tfrac{20}{9} \] \[\frac{x^2}{16/9}-\frac{y^2}{20/9}=1 \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q22 \[a^2+b^2=\tfrac{36}{9}=4=c^2 \quad \text{(check)} \] Answer: \(\displaystyle \dfrac{9x^2}{16}-\dfrac{9y^2}{20}=1\), i.e. \(\displaystyle 45x^2-36y^2=80\)

  3. Exercise 23

    If the lines 2x−3y=5\displaystyle 2 x-3 y=5 and 3x−4y=7\displaystyle 3 x-4 y=7 are the diameters of a circle of area 154\displaystyle 154 square units, then obtain the equation of the circle.

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    NCERT’s answer
    \(\displaystyle x^2+y^2-2 x+2 y=47\)
    Both diameters pass through the centre, so it is their intersection. \[2x-3y=5,\quad 3x-4y=7 \] \[6x-9y=15,\quad 6x-8y=14 \Rightarrow y=-1,\ x=1 \] \[\text{centre } (1,-1) \] \[\pi r^2=154 \Rightarrow r^2=154\cdot\frac{7}{22}=49 \quad \left(\pi=\tfrac{22}{7}\right) \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q23 \[(x-1)^2+(y+1)^2=49 \] Answer: \(\displaystyle (x-1)^2+(y+1)^2=49\), i.e. \(\displaystyle x^2+y^2-2x+2y-47=0\)

  4. Exercise 24

    Find the equation of the circle which passes through the points (2,3)\displaystyle (2,3) and (4,5)\displaystyle (4,5) and the centre lies on the straight line y−4x+3=0\displaystyle y-4 x+3=0.

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    NCERT’s answer
    \(\displaystyle x^2+y^2-4 x-10 y+25=0\)
    The centre \(\displaystyle (h,k)\) lies on the line, so \(\displaystyle k=4h-3\), and it is equidistant from \(\displaystyle (2,3)\) and \(\displaystyle (4,5)\). \[(h-2)^2+(k-3)^2=(h-4)^2+(k-5)^2 \] \[4h+4k=28 \Rightarrow h+k=7 \] \[h+(4h-3)=7 \Rightarrow h=2,\ k=5 \] \[r^2=(2-2)^2+(5-3)^2=4 \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q24 \[(4-2)^2+(5-5)^2=4 \quad \text{(check)} \] Answer: \(\displaystyle (x-2)^2+(y-5)^2=4\), i.e. \(\displaystyle x^2+y^2-4x-10y+25=0\)

  5. Exercise 25

    Find the equation of a circle whose centre is (3,−1)\displaystyle (3,-1) and which cuts off a chord of length 6\displaystyle 6 units on the line 2x−5y+18=0\displaystyle 2 x-5 y+18=0. [Hint: To determine the radius of the circle, find the perpendicular distance from the centre to the given line.]

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    NCERT’s answer
    \(\displaystyle (x-3)^2+(y+1)^2=38\)
    Perpendicular distance from the centre to the line: \[d = \frac{|2(3) - 5(-1) + 18|}{\sqrt{2^2 + 5^2}} = \frac{29}{\sqrt{29}} = \sqrt{29} \] The perpendicular from the centre bisects the chord, so the half-chord is 3. NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q25 \[r^2 = d^2 + 3^2 = 29 + 9 = 38 \] \[(x-3)^2 + (y+1)^2 = 38 \] \[x^2 + y^2 - 6x + 2y - 28 = 0 \] Answer: \(\displaystyle x^2+y^2-6x+2y-28=0\)

  6. Exercise 26

    Find the equation of a circle of radius 5\displaystyle 5 which is touching another circle x2+y2−2x−4y−20=0\displaystyle x^2+y^2-2 x-4 y-20=0 at (5,5)\displaystyle (5, 5).

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    NCERT’s answer
    \(\displaystyle x^2+y^2-18 x-16 y+120=0\)
    Given circle: \[x^2+y^2-2x-4y-20=0 \Rightarrow C_1=(1,2),\quad r_1=\sqrt{1+4+20}=5 \] The new circle has the same radius and touches it at \(\displaystyle P(5,5)\), so \(\displaystyle C_1, P, C_2\) are collinear with \(\displaystyle PC_2=5\). Touching externally, \(\displaystyle P\) is the midpoint of \(\displaystyle C_1C_2\). NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q26 \[C_2 = 2P - C_1 = (10-1,\ 10-2) = (9,8) \] \[C_1C_2 = \sqrt{8^2+6^2} = 10 = r_1 + r_2 \] \[(x-9)^2+(y-8)^2 = 25 \] \[x^2+y^2-18x-16y+120=0 \] The other side of \(\displaystyle P\) gives \(\displaystyle C_2=C_1\), the given circle itself. Answer: \(\displaystyle x^2+y^2-18x-16y+120=0\)

  7. Exercise 27

    Find the equation of a circle passing through the point (7,3)\displaystyle (7,3) having radius 3\displaystyle 3 units and whose centre lies on the line y=x−1\displaystyle y=x-1.

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    NCERT’s answer
    \(\displaystyle x^2+y^2-8 x-6 y+16=0\)
    Centre lies on \(\displaystyle y=x-1\), so take it as \(\displaystyle (h,\ h-1)\). Radius $\displaystyle 3$ and passing through \(\displaystyle (7,3)\): \[(h-7)^2 + (h-1-3)^2 = 3^2 \] \[2h^2 - 22h + 65 = 9 \] \[h^2 - 11h + 28 = 0 \] \[(h-4)(h-7) = 0 \Rightarrow h = 4 \text{ or } 7 \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q27 Centres \(\displaystyle (4,3)\) and \(\displaystyle (7,6)\): \[(x-4)^2+(y-3)^2 = 9 \quad\Rightarrow\quad x^2+y^2-8x-6y+16=0 \] \[(x-7)^2+(y-6)^2 = 9 \quad\Rightarrow\quad x^2+y^2-14x-12y+76=0 \] Answer: \(\displaystyle x^2+y^2-8x-6y+16=0\) or \(\displaystyle x^2+y^2-14x-12y+76=0\)

  8. Exercise 28

    Find the equation of each of the following parabolas
    (a)
    Directrix x=0\displaystyle x=0, focus at (6,0)\displaystyle (6,0)
    (b)
    Vertex at (0,4)\displaystyle (0,4), focus at (0,2)\displaystyle (0,2)
    (c)
    Focus at (−1,−2)\displaystyle (-1,-2), directrix x−2y+3=0\displaystyle x-2 y+3=0

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    NCERT’s answer
    (a)
    \(\displaystyle y^2=12 x-36\),
    (b)
    \(\displaystyle x^2=32-8 y\),
    (c)
    \(\displaystyle 4 x^2+4 x y+y^2+4 x+32 y+16=0\)
    (a) Focus \(\displaystyle (6,0)\), directrix \(\displaystyle x=0\): distance to focus = distance to directrix. \[\sqrt{(x-6)^2+y^2} = |x| \] \[x^2 - 12x + 36 + y^2 = x^2 \] \[y^2 = 12(x-3) \] (b) Axis is \(\displaystyle x=0\); the focus is below the vertex, so the parabola opens downward with \(\displaystyle a = 4-2 = 2\). \[x^2 = -4a(y-4) = -8(y-4) \] \[x^2 + 8y - 32 = 0 \] (c) Focus \(\displaystyle (-1,-2)\), directrix \(\displaystyle x-2y+3=0\): \[\sqrt{(x+1)^2+(y+2)^2} = \frac{|x-2y+3|}{\sqrt{5}} \] \[5\left(x^2+y^2+2x+4y+5\right) = (x-2y+3)^2 \] \[4x^2+4xy+y^2+4x+32y+16 = 0 \] Answer: (a) \(\displaystyle y^2=12x-36\); (b) \(\displaystyle x^2+8y-32=0\); (c) \(\displaystyle 4x^2+4xy+y^2+4x+32y+16=0\)

  9. Exercise 29

    Find the equation of the set of all points the sum of whose distances from the points (3,0)\displaystyle (3,0) and (9,0)\displaystyle (9,0) is 12.

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    NCERT’s answer
    \(\displaystyle 3 x^2+4 y^2-36 x=0\)
    \[PF_1 + PF_2 = 12 > F_1F_2 = 6,\qquad F_1(3,0),\ F_2(9,0) \] So \(\displaystyle P\) traces an ellipse with foci \(\displaystyle F_1, F_2\) and centre their midpoint \(\displaystyle (6,0)\). NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q29 \[2a = 12 \Rightarrow a = 6 \] \[2c = 6 \Rightarrow c = 3 \] \[b^2 = a^2 - c^2 = 36 - 9 = 27 \] \[\frac{(x-6)^2}{36} + \frac{y^2}{27} = 1 \] \[3x^2 + 4y^2 - 36x = 0 \] Answer: \(\displaystyle 3x^2+4y^2-36x=0\), i.e. \(\displaystyle \dfrac{(x-6)^2}{36}+\dfrac{y^2}{27}=1\)

  10. Exercise 30

    Find the equation of the set of all points whose distance from (0,4)\displaystyle (0,4) are 23\displaystyle \frac{2}{3} of their distance from the line y=9\displaystyle y=9.

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    NCERT’s answer
    \(\displaystyle 9 x^2+5 y^2=180\)
    Let \(\displaystyle P(x,y)\); distance to \(\displaystyle (0,4)\) is \(\displaystyle \tfrac23\) of distance to \(\displaystyle y=9\). \[\sqrt{x^2+(y-4)^2} = \tfrac{2}{3}\,|y-9| \] \[9\left(x^2+y^2-8y+16\right) = 4\left(y^2-18y+81\right) \] \[9x^2 + 5y^2 = 180 \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q30 \[\frac{x^2}{20} + \frac{y^2}{36} = 1 \] Answer: \(\displaystyle 9x^2+5y^2=180\), i.e. \(\displaystyle \dfrac{x^2}{20}+\dfrac{y^2}{36}=1\)