12th Grade > Mathematics
ELLIPSE AND HYPERBOLA MCQs
Total Questions : 30
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Answer: Option D. -> [4, 6]
:
D
For the two ellipse to intersect at four different points
1<a2⇒1<(b−5)2⇒b/ϵ[4,6]
:
D
For the two ellipse to intersect at four different points
1<a2⇒1<(b−5)2⇒b/ϵ[4,6]
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
3x−4y+55=12x+5y−4013
⇒ 21x+77y=265
:
D
ransverse axis is the equation of the angle bisector passing containing point (2, 3), which is given by
3x−4y+55=12x+5y−4013
⇒ 21x+77y=265
Answer: Option A. -> x -y + 1 = 0
:
A
Let y = x + c be a tangent to x23−y22=1, then c2=3−2=1 ⇒ c=±1 So, the required tangents are y=x±1
:
A
Let y = x + c be a tangent to x23−y22=1, then c2=3−2=1 ⇒ c=±1 So, the required tangents are y=x±1
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
y−b2a=1e(x−ae)
It passes through´B(0,b) then
b−b2a=−a
a2−b2=−ab
a4e4=a2b2
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
y−b2a=1e(x−ae)
It passes through´B(0,b) then
b−b2a=−a
a2−b2=−ab
a4e4=a2b2
e4+e2=1
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+b2tan2θ∴Minimumvalueofa2cot2θ+b2tan2θis2ab(∵A.M≥G.M)∴MinimumvalueofMN=a+b
:
A
Equation of tangent at (θ) is
xacosθ+ybsinθ=1ItmeetsaxesatM=(acosθ,0)andN=(0,bsinθ)∴MN=√a2sec2θ+b2cosec2θ=√a2+b2+a2cot2θ+b2tan2θ∴Minimumvalueofa2cot2θ+b2tan2θis2ab(∵A.M≥G.M)∴MinimumvalueofMN=a+b
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+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+2y2−5xy+10=0
Answer: Option B. -> 75x - 16y = 418
:
B
Equation of hyperbola is 25x2−16y2=400
⇒ x216−y225=1
Chord is bisected at (6, 2)
∴ 6x16−2y25=6216−2225
⇒ 75x−16y=418
:
B
Equation of hyperbola is 25x2−16y2=400
⇒ x216−y225=1
Chord is bisected at (6, 2)
∴ 6x16−2y25=6216−2225
⇒ 75x−16y=418
Answer: Option D. -> None of these
:
D
The equation of the normal at (asecϕ,btanϕ) to the hyperbola x2a2−y2b2=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)tanϕ−n=(a2+b2)sinϕ−ncosϕ
∴ sinϕ=blamandcosϕ=(a2+b2)l−na
⇒ a2l2−b2m2=(a2+b2)2−n2
:
D
The equation of the normal at (asecϕ,btanϕ) to the hyperbola x2a2−y2b2=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)tanϕ−n=(a2+b2)sinϕ−ncosϕ
∴ sinϕ=blamandcosϕ=(a2+b2)l−na
⇒ a2l2−b2m2=(a2+b2)2−n2
Answer: Option B. -> at 900
:
B
Let the equation to the rectangular hyperbola be
x2−y2=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
x−ysinα=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.
:
B
Let the equation to the rectangular hyperbola be
x2−y2=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
x−ysinα=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.