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12th Grade > Mathematics

ELLIPSE AND HYPERBOLA MCQs

Total Questions : 30 | Page 1 of 3 pages
Question 1. If the ellipse x24+y21=1 meet the ellipse x21+y2a2=1 in four distinct points and a=b210b+25, then the value b does not satisfy
  1.    (−∞,4)
  2.    (4, 6)
  3.    (6,−∞)
  4.    [4, 6]
 Discuss Question
Answer: Option D. -> [4, 6]
:
D
For the two ellipse to intersect at four different points
1<a21<(b5)2b/ϵ[4,6]
Question 2. The area of the parallelogram formed by the tangents at the points whose eccentric angles are θ,θ+π2,θ+π,θ+3π2 on the ellipse x2a2+y2b2=1 is
  1.    ab
  2.    4ab
  3.    3ab
  4.    2ab
 Discuss Question
Answer: Option B. -> 4ab
:
B
The Area Of The Parallelogram Formed By The Tangents At The ...
x2a2+y2b2=1
Area=πab
Let P =(acosθ,bsinθ)
S=(ae,0)
M(h,k) be the midpoint of PS
(h,k)=(ae+acosθ2,bsinθ2)
(hae2)2(a2)2+k2(b2)2=1
Area=πab4
Ratio=14
Question 3. If a hyperbola passes through (2, 3) and has asymptotes 3x - 4y + 5 = 0 and 12x + 5y - 40 = 0, then the equation of its transverse axis is
 
 
  1.    77x - 21y - 265 = 0
  2.    21x - 77y + 265 = 0
  3.    21x - 77y - 265 = 0
  4.    21x + 77y - 265 = 0
 Discuss Question
Answer: Option D. -> 21x + 77y - 265 = 0
:
D
ransverse axis is the equation of the angle bisector passing containing point (2, 3), which is given by
3x4y+55=12x+5y4013
21x+77y=265
Question 4. The equation of a tangent parallel to y = x drawn to x23y22=1, is
  1.     x -y + 1 = 0
  2.    x - y + 2 = 0
  3.    x + y - 1 = 0
  4.    x + y – 2 = 0
 Discuss Question
Answer: Option A. ->  x -y + 1 = 0
:
A
Let y = x + c be a tangent to x23y22=1, then c2=32=1 c=±1 So, the required tangents are y=x±1
Question 5. The normal at an end of a latus rectum of the ellipse x2a2+y2b2=1 passes through an end of the minor  axis if
  1.    e4+e2=1
  2.    e3+e2=1
  3.    e2+e=1
  4.    e3+e=1
 Discuss Question
Answer: Option A. -> e4+e2=1
:
A
Given ellipse equation is x2a2+y2b2=1
Let P(ae,b2a)be one end of latus rectum.
Slope of normal at P(ae,b2a)=1e
Equation of normal is
yb2a=1e(xae)
It passes through´B(0,b) then
bb2a=a
a2b2=ab
a4e4=a2b2
e4+e2=1
Question 6. A tangent to the ellipse x2a2+y2b2=1 cuts the axes in M and N. Then the least length of MN is
  1.    a + b
  2.    a - b
  3.    a2+b2
  4.    a2−b2
 Discuss Question
Answer: Option A. -> a + b
:
A
Equation of tangent at (θ) is
xacosθ+ybsinθ=1ItmeetsaxesatM=(acosθ,0)andN=(0,bsinθ)MN=a2sec2θ+b2cosec2θ=a2+b2+a2cot2θ+b2tan2θMinimumvalueofa2cot2θ+b2tan2θis2ab(A.MG.M)MinimumvalueofMN=a+b
Question 7. If foci of hyperbola lie on y = x and one of the asymptotes is y = 2x, then equation of the hyperbola, given  that it passes through (3, 4), is
  1.    x2−y2−52xy+5=0
  2.    2x2−2y2+5xy+5=0
  3.    2x2+2y2−5xy+10=0
  4.     None of these
 Discuss Question
Answer: Option C. -> 2x2+2y2−5xy+10=0
:
C
Foci of hyperbola lie on y = x. So, the major axis is y = x. Major axis of hyperbola bisects the asymptote.
Equation of hyperbola is x = 2y
Equation of hyperbola is (y – 2x)(x – 2y) + k = 0
Given that, it passes through (3, 4)
Hence, required equation is
2x2+2y25xy+10=0
Question 8. Equation of the chord of the hyperbola 25x216y2=400 which is bisected at the point (6, 2), is
 
  1.    16x – 75y = 418
  2.    75x - 16y = 418
  3.    25x - 4y = 400
  4.    None of these
 Discuss Question
Answer: Option B. -> 75x - 16y = 418
:
B
Equation of hyperbola is 25x216y2=400
x216y225=1
Chord is bisected at (6, 2)
6x162y25=62162225
75x16y=418
Question 9. The line lx + my + n = 0 will be a normal to the hyperbola b2x2a2y2=a2b2, if
  1.    a2l2+b2m2=(a2+b2)2n2  
  2.    a2l2+b2m2=(a2−b2)2n2  
  3.    a2l2+b2m2=(a2+b2)2n
  4.    None of these
 Discuss Question
Answer: Option D. -> None of these
:
D
The equation of the normal at (asecϕ,btanϕ) to the hyperbola x2a2y2b2=1 is
ax sinϕ+by=(a2+b2)tanϕ ......(i)
and the equation of the line is
lx + my + n =0 …… (ii)
now, equations (i) and (ii) represent the same line, therefore
2sinϕl=bm=(a2+b2)tanϕn=(a2+b2)sinϕncosϕ
sinϕ=blamandcosϕ=(a2+b2)lna
a2l2b2m2=(a2+b2)2n2
Question 10. Two concentric hyperbolas,whose axes meet at angle of 45,cut
  1.    at 450
  2.    at 900
  3.    Nothing can be said       
  4.    None of these
 Discuss Question
Answer: Option B. -> at 900
:
B
Let the equation to the rectangular hyperbola be
x2y2=a2 …… (i)
As the asymptotes of this are the axes of the other and vice-versa, hence the equation of the other hyperbola may be written as
xy=c2 …… (ii)
Let equations (i) and (ii) meet at some point whose coordinates are
(asecα,atanα)
Then, the tangent at the point a (secα,atanα) to equation (ii) is
xysinα=acosα …… (iii)
and the tangent at the point (asecα,atanα) to equation (ii) is
y+xsinα=(2c2a)cosα …… (iv)
clearly, the slopes of the tangents given by equations (iv) and (iii) are respectively - secαand1sinα , so their product is secα.(1sinα)=1 Hence, the tangents are at right angle.

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