Free Objective Test 02 Practice Test - 11th and 12th
Question 1
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
SOLUTION
Solution : B
x2a2+y2b2=1
Area=πab
Let P =(a cosθ,b sinθ)
S=(ae,0)
M(h,k) be the midpoint of PS
(h,k)=(ae+a cosθ2,b sinθ2)
(h−ae2)2(a2)2+k2(b2)2=1
Area=πab4
Ratio=14
Question 2
If the ellipse x24+y21=1 meet the ellipse x21+y2a2=1 in four distinct points and a=b2−10b+25, then the value b does not satisfy
SOLUTION
Solution : D
For the two ellipse to intersect at four different points
1<a2⇒1<(b−5)2⇒b/ϵ[4,6]
Question 3
If the ellipse x2a2−3+y2a+4=1 is inscribed in a square of side length a√2 then a is
SOLUTION
Solution : D
Sides of the square will be perpendicular tangents to the ellipse so, vertices of the square will lie on director circle. So diameter of director circle is
2√(a2−3)+(a+4)=√2a2+2a22√a2+a+1=2a⇒a=−1But for ellipse a2>3&a>−4
So there is no value of a which satisfy both.
Question 4
A man running round a race course notes that the sum of the distances of two flag posts from him is 8 meters.The area of the path he encloses in square meters if the distance between flag posts is 4 is
SOLUTION
Solution : D
The path he runs is an ellipse whose major axis is 4m,e=12
Area=π ab
Question 5
A tangent to the ellipse x2a2+y2b2=1 cuts the axes in M and N. Then the least length of MN is
SOLUTION
Solution : A
Equation of tangent at (θ) is
xacosθ+ybsinθ=1It meets axes at M=(acosθ,0)and N=(0,bsin θ)∴MN=√a2sec2θ+b2cosec2θ=√a2+b2+a2cot2θ+b2tan2θ∴Minimum value of a2cot2θ+b2tan2θ is 2ab(∵A.M≥G.M)∴Minimum value of MN=a+b
Question 6
Tangents are drawn from any point on the circle x2+y2=41 to the Ellipse x225+y216=1 then the angle between the two tangents is
SOLUTION
Solution : D
The given circle is the director circle of the ellipse,so the angle between the tangents is π2.
Question 7
The perimeter of a triangle is 20 and the points (-2, -3) and (-2, 3) are two of the vertices of it. Then the locus of third vertex is :
SOLUTION
Solution : C
A(−2,−3),B(−2,3)
Let the third vertex be C(x,y)
AB + BC+CA=20
BC+CA=14
So locus of C is ellipse with A,B as focii.
2a=14
2ae=6
e=37
∴ we get b as √40
∴ Equation of ellipse is (x+2)240+y249=1
Question 8
The line lx + my + n = 0 will be a normal to the hyperbola b2x2−a2y2=a2b2, if
SOLUTION
Solution : 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=blamand cosϕ=(a2+b2)l−na
⇒ a2l2−b2m2=(a2+b2)2−n2
Question 9
Equation of the chord of the hyperbola 25x2−16y2=400 which is bisected at the point (6, 2), is
SOLUTION
Solution : B
Equation of hyperbola is 25x2−16y2=400
⇒ x216−y225=1
Chord is bisected at (6, 2)
∴ 6x16−2y25=6216−2225
⇒ 75x−16y=418
Question 10
The equation of a hyperbola, conjugate to the hyperbola x2+3xy+2y2+2x+3y+1=0, is
SOLUTION
Solution : B
We know that,
Equation of hyperbola + Equation of conjugate
hyperbola = 2(Equation of asymptotes)
∴ Equation of conjugate hyperbola
= 2 Equation of asymptotes - Equation of hyperbola
C(x, y) = 2A(x, y) – H(x, y) …… (i)
∵ H(x,y)≡x2+3xy+2y2+2x+3y=0
∴ Equation of asymptotes is
x2+3xy+2y2+2x+3y+λ=0
∴ Δ=0,abc+2fgh−af2−bg2−ch2=0,thenλ=1
∴ A(x,y)≡x2+3xy+2y2+2x+3y+1=0
∴ C(x,y)≡x2+3xy+2y2+2x+3y+2=0[from equation (i)]
Question 11
The eccentricity of the rectangular hyperbola is
SOLUTION
Solution : B
In case of rectangular hyperbola,
a=b⇒b2=a2⇒a2(e2−1)=a2⇒e=√2
Question 12
Two concentric hyperbolas,whose axes meet at angle of 45∘,cut
SOLUTION
Solution : 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
(a secα,a tanα)
Then, the tangent at the point a (secα,atanα) to equation (ii) is
x−ysinα=acosα …… (iii)
and the tangent at the point (a secα,atanα) to equation (ii) is
y+xsinα=(2c2a)cosα …… (iv)
clearly, the slopes of the tangents given by equations (iv) and (iii) are respectively - secα and1sinα , so their product is −secα.(1sinα)=−1 Hence, the tangents are at right angle.
Question 13
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
SOLUTION
Solution : 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
Question 14
If two tangents can be drawn to the different branches of hyperbola
x21−y24=1 from the paint (α,α2), then
SOLUTION
Solution : C
Given that,
x21−y24=1
Since, (α,α2) lie on the parabola y=x2, then (α,α2) must lie between the asymptotes of hyperbola x21−y24=1 in 1st and 2nd quadrants.
So, the asymptotes are y=±2x
∴ 2α<α2
⇒ α<0orα>2 and −2α<α2
α<−2or−2α<0
∴ αϵ(−∞,−2)or(2,∞)
Question 15
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
SOLUTION
Solution : 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