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NCERT Exemplar · Class 11 Mathematics Conic Sections

59 questions · 59 still being checked

EXERCISE 11.3 11–20 (part 2 of 6)

  1. Exercise 11

    If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

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    NCERT’s answer
    \(\displaystyle \frac{\sqrt{3}}{2}\)
    For \(\displaystyle \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) with \(\displaystyle a > b\): \[\text{latus rectum} = \frac{2b^2}{a}, \qquad \text{minor axis} = 2b \] \[\frac{2b^2}{a} = \frac{1}{2}(2b) = b \] \[a = 2b \] \[e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{1}{4} = \frac{3}{4} \] Answer: \(\displaystyle e = \dfrac{\sqrt{3}}{2}\)

  2. Exercise 12

    Given the ellipse with equation 9x2+25y2=225\displaystyle 9 x^2+25 y^2=225, find the eccentricity and foci.

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    NCERT’s answer
    ecentricity \(\displaystyle =\frac{4}{5}\) and foci \(\displaystyle (4,0)\) and \(\displaystyle (-4,0)\)
    \[9x^2 + 25y^2 = 225 \] \[\frac{x^2}{25} + \frac{y^2}{9} = 1 \] \[a^2 = 25,\quad b^2 = 9 \] \[c^2 = a^2 - b^2 = 16 \Rightarrow c = 4 \] \[e = \frac{c}{a} = \frac{4}{5} \] Foci lie on the major (x) axis at \(\displaystyle (\pm c, 0)\). Answer: \(\displaystyle e = \dfrac{4}{5}\), foci \(\displaystyle (\pm 4, 0)\)

  3. Exercise 13

    If the eccentricity of an ellipse is 58\displaystyle \frac{5}{8} and the distance between its foci is 10\displaystyle 10, then find latus rectum of the ellipse.

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    NCERT’s answer
    \(\displaystyle \frac{39}{4}\)
    \[2ae = 10,\qquad e = \frac{5}{8} \] \[2a \cdot \frac{5}{8} = 10 \Rightarrow a = 8 \] \[b^2 = a^2(1 - e^2) = 64\left(1 - \frac{25}{64}\right) = 39 \] \[\text{latus rectum} = \frac{2b^2}{a} = \frac{2 \cdot 39}{8} \] Answer: \(\displaystyle \dfrac{39}{4}\)

  4. Exercise 14

    Find the equation of ellipse whose eccentricity is 23\displaystyle \frac{2}{3}, latus rectum is 5\displaystyle 5 and the centre is (0,0)\displaystyle (0,0).

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    NCERT’s answer
    \(\displaystyle \frac{4 x^2}{81}+\frac{4 y^2}{45}=1\)
    Take the foci on the x-axis: \(\displaystyle \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\). \[b^2 = a^2(1 - e^2) = a^2\left(1 - \frac{4}{9}\right) = \frac{5a^2}{9} \] \[\frac{2b^2}{a} = 5 \Rightarrow \frac{10a}{9} = 5 \Rightarrow a = \frac{9}{2} \] \[a^2 = \frac{81}{4},\qquad b^2 = \frac{5}{9}\cdot\frac{81}{4} = \frac{45}{4} \] \[\frac{x^2}{81/4} + \frac{y^2}{45/4} = 1 \] If the major axis is along the y-axis, interchange \(\displaystyle x\) and \(\displaystyle y\). Answer: \(\displaystyle \dfrac{4x^2}{81} + \dfrac{4y^2}{45} = 1\)

  5. Exercise 15

    Find the distance between the directrices of the ellipse x236+y220=1\displaystyle \frac{x^2}{36}+\frac{y^2}{20}=1.

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    NCERT’s answer
    $\displaystyle 18$
    \[a^2 = 36,\quad b^2 = 20 \] \[c^2 = a^2 - b^2 = 16 \Rightarrow c = 4 \] \[e = \frac{c}{a} = \frac{4}{6} = \frac{2}{3} \] Directrices: \[x = \pm \frac{a}{e} = \pm \frac{6}{2/3} = \pm 9 \] \[\text{distance} = 9 - (-9) \] Answer: \(\displaystyle 18\)

  6. Exercise 16

    Find the coordinates of a point on the parabola y2=8x\displaystyle y^2=8 x whose focal distance is 4.

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    NCERT’s answer
    $\displaystyle (2, 4)$ , $\displaystyle (2, -4)$
    \[y^2 = 4ax = 8x \Rightarrow a = 2 \] Focal distance of \(\displaystyle (x_1, y_1)\) is \(\displaystyle x_1 + a\): \[x_1 + 2 = 4 \Rightarrow x_1 = 2 \] \[y_1^2 = 8 \cdot 2 = 16 \Rightarrow y_1 = \pm 4 \] Answer: \(\displaystyle (2, 4)\) and \(\displaystyle (2, -4)\)

  7. Exercise 17

    Find the length of the line-segment joining the vertex of the parabola y2=4ax\displaystyle y^2=4 a x and a point on the parabola where the line-segment makes an angle θ\displaystyle \theta to the x\displaystyle x-axis.

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    NCERT’s answer
    \(\displaystyle \frac{4 a \cos \theta}{\sin ^2 \theta}\)
    Take the vertex \(\displaystyle O\) as origin and \(\displaystyle OP=r\). NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q17 \[P=(r\cos\theta,\ r\sin\theta) \] \[r^2\sin^2\theta = 4a\,r\cos\theta \quad \text{(} P \text{ lies on } y^2=4ax \text{)} \] \[r\neq 0 \Rightarrow r=\frac{4a\cos\theta}{\sin^2\theta} \] Answer: \(\displaystyle OP=\dfrac{4a\cos\theta}{\sin^2\theta}=4a\cot\theta\csc\theta\)

  8. Exercise 18

    If the points (0,4)\displaystyle (0,4) and (0,2)\displaystyle (0,2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

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    NCERT’s answer
    \(\displaystyle x^2+8 y=32\)
    The axis is the \(\displaystyle y\)-axis and the focus lies below the vertex, so the parabola opens downward. \[a = 4-2 = 2 \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q18 \[x^2 = -4a\,(y-4) \] \[x^2 = -8\,(y-4) \] Answer: \(\displaystyle x^2=-8(y-4)\), i.e. \(\displaystyle x^2+8y-32=0\)

  9. Exercise 19

    If the line y=mx+1\displaystyle y=m x+1 is tangent to the parabola y2=4x\displaystyle y^2=4 x then find the value of m\displaystyle m. [Hint: Solving the equation of line and parabola, we obtain a quadratic equation and then apply the tangency condition giving the value of m\displaystyle m].

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    NCERT’s answer
    \(\displaystyle m=1\)
    Substitute \(\displaystyle y=mx+1\) in \(\displaystyle y^2=4x\): \[(mx+1)^2 = 4x \] \[m^2x^2+(2m-4)x+1=0 \] For \(\displaystyle m\neq 0\) this is a genuine quadratic; tangency means equal roots, \(\displaystyle D=0\). \[(2m-4)^2-4m^2=0 \] \[16-16m=0 \] \[m=1 \] \[(x+1)^2=4x \Rightarrow (x-1)^2=0 \quad \text{(touches at } (1,2)\text{)} \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q19 Answer: \(\displaystyle m=1\)

  10. Exercise 20

    If the distance between the foci of a hyperbola is 16\displaystyle 16 and its eccentricity is 2\displaystyle \sqrt{2}, then obtain the equation of the hyperbola.

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    NCERT’s answer
    \(\displaystyle x^2-y^2=32\)
    Take the centre at the origin and the foci on the \(\displaystyle x\)-axis (on the \(\displaystyle y\)-axis the same working gives \(\displaystyle y^2-x^2=32\)). \[2ae=16,\quad e=\sqrt2 \Rightarrow a=\frac{8}{\sqrt2}=4\sqrt2,\quad a^2=32 \] \[b^2=a^2(e^2-1)=32(2-1)=32 \] \[\frac{x^2}{32}-\frac{y^2}{32}=1 \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q20 Answer: \(\displaystyle \dfrac{x^2}{32}-\dfrac{y^2}{32}=1\), i.e. \(\displaystyle x^2-y^2=32\)