Conic Sections - Test Papers

 CBSE Test Paper 01

CH-11 Conic Sections

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1. The equation of the tangent to the conic x2−y2−8x+2y+11=0$x^2-y^2-8x+2y+11=0$ at (2, 1) is

   1. 2 x + 1 = 0

   2. x – 2 = 0

   3. x +2 = 0

   4. x + y + 1 = 0

2. The equation(x2+y2)+5x−7y−2=0$(x^2+y^2)+5x-7y-2=0$ represents

   1. a circle

   2. an empty set

   3. a degenerate circle

   4. a pair of straight lines

3. Three normals to the parabola y2=x$y^2=x$ are drawn through a point (c, 0) then

   1. none of these

   2. c>12$c>\frac{1}{2}$

   3. c=12$c=\frac{1}{2}$

   4. c=14$c=\frac{1}{4}$

4. The graph of the function f(x) = 1xi.e.$\frac{1}{x}i.e.$ the curve y=1x$y=\frac{1}{x}$ is

   1. a hyperbola

   2. a parabola

   3. an ellipse

   4. a circle

5. The ellipse = x2a2+y2b2=1,b2=a2$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,b^2=a^2$ is a

   1. a hyperbola

   2. none of these.

   3. horizontal ellipse

   4. vertical ellipse

6. Fill in the blanks:

   The equation of the circle having centre at (3, -4) and touching the line 5x + 12y - 12 = 0 is ________.

7. Fill in the blanks:

   ________ of the hyperbola is the ratio of the distance of any one focus from the centre and the distance of any one vertex from the centre.

8. Find the equation of parabola when the vertex is at (0, 0) and focus is at (0, 4).

9. What is the condition that the equation, on comparing with general equation of circle, ax2 + by2 + 6x  + 3y + hxy + 3 = 0 is the equation of circle?

10. Find the equation of hyperbola having Foci (0, ±13) and the conjugate axis is of length 24.

11. Determine whether x2 + y2 + 2x  + 10y + 26 = 0 represent a circle or point.

12. Find the equation of ellipse having Major axis on the x-axis and passes through the points (4, 3) and (6, 2)

13. Find the equation of ellipse having Length of minor axis 16, foci (0,±$\pm$6)

14. Find the centre and radius of the circle. x 2 + y 2 - 8x - 10y - 12 = 0

15. Find the equation of the hyperbola whose foci are (4, 2) and (8,2) and eccentricity is 2.

CBSE Test Paper 01
CH-11 Conic Sections

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Solution

1. (b) x – 2 = 0
   Explanation: Differentiating the given equation w.r.t x, we get, 2x -2ydydx$\frac{dy}{dx}$ -8 + 2 dydx$\frac{dy}{dx}$ = 0 dydx$\frac{dy}{dx}$(1-y) = x - 4 Therefore dydx$\frac{dy}{dx}$ = x−41−y$\frac{x-4}{1-y}$ Therefore dydx$\frac{dy}{dx}$(2,1) is not defined  The equation of the tangent at (x1, y1) is y-y1 = m(x - x1) Therefore the equation of the tangent is x-2 = 0
2. (a) a circle
   Explanation: The general equation of the circle is x2+y2-2gh-2fy+c = 0. Sice the given equations satisfies the general equation, it represents the equation of the circle.
3. (b) c>12$c>\frac{1}{2}$ Explanation: The equation of the normal to a parabola y2 = 4ax is y = mx -2am - am3 Hence the equation of the normal to the given parabola y2 = x is  mx - m2$\frac{m}{2}$- m34$\frac{m^3}{4}$ Since it passes throught (c,0) mc -m2$\frac{m}{2}$- m34$\frac{m^3}{4}$ = 0 on solving we  get  m = 0 or m2 = 4(c-1/2) If m = 0 then the equation of the normal is y = 0 If m2 ≥$\ge$0, then  4(c-1/2) ≥$\ge$0 Hence c-1/2 ≥$\ge$0 or c > 1/2
4. (a) a hyperbola
   Explanation: it is called rectangular hyperbola.
5. (b) none of these.
   Explanation: If a2 = b2, then the equation becomes x2 + y2 = a2 which represents the equation of a circle.
6. (x - 3)2 + (y + 4)2 = (4513)2${\left(\frac{45}{13}\right)}^2$
7. Eccentricity
8. Since, the vertex is at (0, 0) and focus is at (0, 4) which lies on Y-axis. The Y-axis is the axis of the parabola.
   ∴$\therefore$ Equation of parabola is of the form
   x2 = - 4ay ⇒$\Rightarrow$ x2 = - 4(4)y [∵$\because$ a = 4]
   ⇒$\Rightarrow$ x2 = - 16y
9. Given, equation will represent a circle, if Coefficient of x2 = Coefficient of y2
   i.e., a = b and coefficient of xy should be zero.
   i.e., h = 0.
10. Here foci are (0, ±13) which lie on y-axis.
    So the equation of hyperbola in standard form is y2a2−x2b2=1$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
    ∴$\therefore$ (13)2 = a2 + (12)2 ⇒ a2 = 169 – 144 = 25
    Thus required equation of hyperbola is
    y225−x2(12)2=1⇒y225−x2144=1$\frac{y^2}{25}-\frac{x^2}{{(12)}^2}=1\Rightarrow \frac{y^2}{25}-\frac{x^2}{144}=1$
11. We have,  x2 + y2 + 2x  + 10y + 26 = 0
    On adding 1 and 25 both sides to make perfect squares, we get
    (x2 + 2x + 1) +  (y2  +  l0y +  25) = - 26 + 1 + 25
    ⇒$\Rightarrow$ (x + 1)2 + (y + 5)2 = 
    ⇒$\Rightarrow$ [x - (- 1)]2 + [y - (- 5)]2 = 02
    Hence, it represents a point circle, because it has zero radius.

12. Since the major axis is along x-axis.
    So the equation of ellipse in standard form is x2a2+y2b2=1$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
    Since the ellipse passes through point (4, 3)
    ∴16a2+9b2=1$\therefore \frac{16}{a^2}+\frac{9}{b^2}=1$. . . (i)
    Also the ellipse passes through point (6, 2)
    ∴36a2+4b2=1$\therefore \frac{36}{a^2}+\frac{4}{b^2}=1$....(ii)
    Solving (i) and (ii), we have
    a2 = 52 and b2 = 13
    Thus equation of required ellipse is
    x252+y213=1$\frac{x^2}{52}+\frac{y^2}{13}=1$

13. The foci (0,±$\pm$6) lie on y-axis.
    So the equation of ellipse in standard form is x2b2+y2a2=1]$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1]$
    Now length of minor axis 2b = 16 ⇒$\Rightarrow$ b = 8
    foci (0, ±c) is (0,± 6) ⇒ c = 6
    We know that c2=a2−b2$c^2=a^2-b^2$
    ∴$\therefore$ (6)2 = a2 - (8)2 ⇒$\Rightarrow$ a2 = 36 + 64 = 100
    Thus equation of required ellipse is
    x264+y2100=1$\frac{x^2}{64}+\frac{y^2}{100}=1$

14. The given equation of circle is
    x 2 +y 2  - 8x - 10y - 12 = 0
    ∴$\therefore$ (x 2 - 8x) + (y 2 + 10y) = 12
    Completing the square
    ⇒$\Rightarrow$ [x 2 - 8x + (4) 2] + [y 2 + 10y + (5) 2]
    = 12 + (4) 2  + (5) 2
    ⇒$\Rightarrow$ (x - 4) 2 + (y + 5) 2 = 12 + 16 + 25
    ⇒$\Rightarrow$ (x - 4) 2 + (y + 5) 2 = 53
    ⇒$\Rightarrow$ (x - 4) 2 + (y + 5) 2 = (53−−√)2$(\sqrt{53}{)}^2$
    Comparing it with (x - h) 2 + (y - k) 2 = r 2, we have
    h = 4, k = -5 and  r = 53−−√$\sqrt{53}$
    Thus coordinates of the centre is (4, -5) and radius is 53−−√$\sqrt{53}$.

15. The centre of the hyperbola is the mid-point of the line joining the two foci.
    So, the coordinates of the centre are (4+82,2+22)$\left(\frac{4+8}{2},\frac{2+2}{2}\right)$ i.e., (6, 2).
    Let 2a and 2b be the length of transverse and conjugate axes and let e be the eccentricity.
    Then, the equation of the hyperbola is
    (x−6)2a2−(y−2)2b2$\frac{(x-6{)}^2}{a^2}-\frac{(y-2{)}^2}{b^2}$ = 1  ...(i)
    Now, the distance between two foci = 2ae
    ⇒$\Rightarrow$ (8−4)2+(2−2)2−−−−−−−−−−−−−−−√$\sqrt{(8-4{)}^2+(2-2{)}^2}$ = 2ae [∵$\because$ foci = (4, 2) and (8, 2)]
    ⇒$\Rightarrow$ (4)2−−−√$\sqrt{(4{)}^2}$ = 2ae
    ⇒$\Rightarrow$ 2ae = 4
    ⇒$\Rightarrow$ 2 ×$\times$ a ×$\times$ 2 = 4 [∵$\because$ e = 2]
    ⇒$\Rightarrow$ a = 44$\frac{4}{4}$ = 1
    ⇒$\Rightarrow$ a2 = 1
    Now,
    b2 = a2 (e2 - 1)
    ⇒$\Rightarrow$ b2 = 1 (22 - 1) [∵$\because$ e = 2]
    ⇒$\Rightarrow$ b2 = 4 - 1
    ⇒$\Rightarrow$ b2 = 3
    Putting a2 = 1 and b2 = 3 in equation (i), we get
    (x−6)21−(y−2)23$\frac{(x-6{)}^2}{1}-\frac{(y-2{)}^2}{3}$ = 1
    ⇒$\Rightarrow$ 3(x−6)2−(y−2)23$\frac{3(x-6{)}^2-(y-2{)}^2}{3}$ = 1
    ⇒$\Rightarrow$ 3 (x - 6)2 - (y - 2)2 = 3
    ⇒$\Rightarrow$ 3[x2 + 36 − 12x] − [y2 + 4 − 4y] = 3
    ⇒$\Rightarrow$ 3x2 + 108 − 36x − y2 − 4 + 4y = 3
    ⇒$\Rightarrow$ 3x2 − y2 − 36x + 4y + 101 = 0
    This is the equation of the required hyperbola.