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NCERT Exemplar · Class 11 Mathematics Conic Sections

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EXERCISE 11.3 31–40 (part 4 of 6)

  1. Exercise 31

    Show that the set of all points such that the difference of their distances from (4,0)\displaystyle (4,0) and (−4,0)\displaystyle (- 4, 0) is always equal to 2\displaystyle 2 represent a hyperbola.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle P(x,y)\), \(\displaystyle F_1(4,0)\), \(\displaystyle F_2(-4,0)\), with \(\displaystyle PF_2 - PF_1 = 2\). The case \(\displaystyle PF_1 - PF_2 = 2\) gives the same equation, by symmetry. \[d_1 = \sqrt{(x-4)^2+y^2},\qquad d_2 = \sqrt{(x+4)^2+y^2} \] \[d_2 - d_1 = 2 \] \[d_2^2 - d_1^2 = 16x \Rightarrow d_2 + d_1 = 8x \] \[d_2 = 4x + 1 \] \[(x+4)^2 + y^2 = (4x+1)^2 \] \[15x^2 - y^2 = 15 \] NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q31 \[x^2 - \frac{y^2}{15} = 1 \] This is \(\displaystyle \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) with \(\displaystyle a=1,\ b^2=15\); then \(\displaystyle c^2=a^2+b^2=16\), so the foci are \(\displaystyle (\pm4,0)\). Every point of it has \(\displaystyle |PF_1-PF_2|=2a=2\). Answer: the locus is the hyperbola \(\displaystyle x^2-\dfrac{y^2}{15}=1\).

  2. Exercise 32

    Find the equation of the hyperbola with
    (a)
    Vertices (±5,0)\displaystyle ( \pm 5,0), foci (±7,0)\displaystyle ( \pm 7,0)
    (b)
    Vertices (0,±7),e=43\displaystyle (0, \pm 7), e=\frac{4}{3}
    (c)
    Foci (0,±10)\displaystyle (0, \pm \sqrt{10}), passing through (2,3)\displaystyle (2,3)

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    (a) Transverse axis along the \(\displaystyle x\)-axis: \(\displaystyle a=5,\ c=7\). \[b^2 = c^2 - a^2 = 49 - 25 = 24 \] \[\frac{x^2}{25} - \frac{y^2}{24} = 1 \] (b) Transverse axis along the \(\displaystyle y\)-axis: \(\displaystyle a=7,\ e=\tfrac43\). \[b^2 = a^2(e^2-1) = 49\left(\tfrac{16}{9}-1\right) = \tfrac{343}{9} \] \[\frac{y^2}{49} - \frac{9x^2}{343} = 1 \] (c) Foci on the \(\displaystyle y\)-axis, \(\displaystyle c^2 = 10\): \(\displaystyle \dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1\) with \(\displaystyle b^2 = 10 - a^2\). Through \(\displaystyle (2,3)\): \[\frac{9}{a^2} - \frac{4}{10-a^2} = 1 \] \[a^4 - 23a^2 + 90 = 0 \] \[(a^2-5)(a^2-18) = 0 \Rightarrow a^2 = 5 \quad (a^2<10) \] \[b^2 = 5,\qquad \frac{y^2}{5} - \frac{x^2}{5} = 1 \] Answer: (a) \(\displaystyle \dfrac{x^2}{25}-\dfrac{y^2}{24}=1\); (b) \(\displaystyle \dfrac{y^2}{49}-\dfrac{9x^2}{343}=1\); (c) \(\displaystyle y^2-x^2=5\)
  3. State Whether the statements in each of the Exercises from $\displaystyle 33$ to $\displaystyle 40$ are True or False. Justify

    Exercise 33

    The line x+3y=0\displaystyle x+3 y=0 is a diameter of the circle x2+y2+6x+2y=0\displaystyle x^2+y^2+6 x+2 y=0.

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    NCERT’s answer
    False
    False. A diameter passes through the centre.\[x^2+y^2+6x+2y=0 \;\Rightarrow\; \text{centre } (-3,-1) \]\[x+3y = -3+3(-1) = -6 \neq 0 \]The centre is not on the line, so it is not a diameter.

  4. Exercise 34

    The shortest distance from the point (2,−7)\displaystyle (2, -7) to the circle x2+y2−14x−10y−151=0\displaystyle x^2+y^2-14 x-10 y-151=0 is equal to 5. [Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point.]

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    NCERT’s answer
    False
    False. The point is inside the circle, so the shortest distance is \(\displaystyle r-CP\).\[(x-7)^2+(y-5)^2 = 49+25+151 = 225 \;\Rightarrow\; C(7,5),\; r=15 \]\[CP=\sqrt{(2-7)^2+(-7-5)^2}=\sqrt{25+144}=13 \]\[r-CP = 15-13 = 2 \neq 5 \]

  5. Exercise 35

    If the line lx+my=1\displaystyle l x+m y=1 is a tangent to the circle x2+y2=a2\displaystyle x^2+y^2=a^2, then the point (l,m)\displaystyle (l, m) lies on a circle. [Hint: Use that distance from the centre of the circle to the given line is equal to radius of the circle.]

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    NCERT’s answer
    True
    True. Tangency means the distance from the centre \(\displaystyle (0,0)\) to the line equals the radius.\[\frac{|\,l\cdot 0+m\cdot 0-1\,|}{\sqrt{l^2+m^2}} = a \]\[l^2+m^2=\frac{1}{a^2} \]So \(\displaystyle (l,m)\) lies on the circle \(\displaystyle x^2+y^2=\dfrac{1}{a^2}\).

  6. Exercise 36

    The point (1,2)\displaystyle (1,2) lies inside the circle x2+y2−2x+6y+1=0\displaystyle x^2+y^2-2 x+6 y+1=0.

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    NCERT’s answer
    False
    False. Evaluate the circle's expression at the point.\[S_1 = 1^2+2^2-2(1)+6(2)+1 = 16 > 0 \]Positive means the point lies outside the circle.Check: centre \(\displaystyle (1,-3)\), \(\displaystyle r=3\), and\[\sqrt{(1-1)^2+(2+3)^2}=5>3 \]

  7. Exercise 37

    The line lx+my+n=0\displaystyle l x+m y+n=0 will touch the parabola y2=4ax\displaystyle y^2=4 a x if ln=am2\displaystyle l n=a m^2.

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    NCERT’s answer
    True
    True. Put \(\displaystyle x=-\dfrac{my+n}{l}\) (\(\displaystyle l\neq 0\)) into \(\displaystyle y^2=4ax\).\[y^2=-\frac{4a(my+n)}{l} \]\[l\,y^2+4am\,y+4an=0 \]Tangency needs a zero discriminant.\[(4am)^2-4\,l\,(4an)=0 \]\[16a\,(am^2-ln)=0 \;\Rightarrow\; ln=am^2 \]

  8. Exercise 38

    If P is a point on the ellipse x216+y225=1\displaystyle \frac{x^2}{16}+\frac{y^2}{25}=1 whose foci are S and S′\displaystyle \mathrm{S}^{\prime}, then PS+PS′=8\displaystyle \mathrm{PS}+\mathrm{PS}^{\prime}=8.

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    NCERT’s answer
    False
    False. Here \(\displaystyle 25>16\), so the major axis lies along the \(\displaystyle y\)-axis and the semi-major axis is \(\displaystyle \sqrt{25}=5\).\[PS+PS' = 2\cdot(\text{semi-major axis}) \]\[PS+PS' = 2\cdot 5 = 10 \neq 8 \]

  9. Exercise 39

    The line 2x+3y=12\displaystyle 2 x+3 y=12 touches the ellipse x29+y24=2\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=2 at the point (3,2)\displaystyle (3,2).

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    NCERT’s answer
    True
    True. The point lies on the ellipse, and the line is the tangent there.\[\frac{3^2}{9}+\frac{2^2}{4}=1+1=2 \]Tangent at \(\displaystyle (x_1,y_1)\) is \(\displaystyle \dfrac{xx_1}{9}+\dfrac{yy_1}{4}=2\).\[\frac{3x}{9}+\frac{2y}{4}=2 \]\[2x+3y=12 \]

  10. Exercise 40

    The locus of the point of intersection of lines 3x−y−43k=0\displaystyle \sqrt{3} x-y-4 \sqrt{3} k=0 and 3kx+ky−43=0\displaystyle \sqrt{3} k x+k y-4 \sqrt{3}=0 for different value of k\displaystyle k is a hyperbola whose eccentricity is 2. [Hint:Eliminate k\displaystyle k between the given equations]

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    NCERT’s answer
    True
    True. Solve each line for \(\displaystyle k\) and equate.\[k=\frac{\sqrt3\,x-y}{4\sqrt3},\qquad k=\frac{4\sqrt3}{\sqrt3\,x+y} \]\[(\sqrt3\,x-y)(\sqrt3\,x+y)=(4\sqrt3)(4\sqrt3)=48 \]\[3x^2-y^2=48 \;\Rightarrow\; \frac{x^2}{16}-\frac{y^2}{48}=1 \]\[e=\sqrt{1+\frac{48}{16}}=\sqrt4=2 \]