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Conic Sections
Quiz 1
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The locus of the point from which the tangent to the circles x² + y² - 4 = 0 and x² + y² - 8x + 15 = 0 are equal is given by the equation
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8x + 19 = 0
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8x - 19 = 0
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4x - 19 = 0
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4x + 19 = 0
Explanation
8x - 19 = 0
Hint:
Given equation of circles are x² + y² - 4 = 0 and x² + y² - 8x + 15 = 0
Now, the required line is the radical axis of the two circles are
(x² + y² - 4) - (x² + y² - 8x + 15) = 0
⇒ x² + y² - 4 - x² - y² + 8x - 15 = 0
⇒ 8x - 19 = 0
The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0
0%
7
0%
8
0%
9
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10
Explanation
7
Hint:
The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)
= {9 + 16 + 10}/√(9 + 16)
= 35/√25
= 35/5
= 7
A man running a race course notes that the sum of the distances from the two flag posts from him is always 10 meter and the distance between the flag posts is 8 meter. The equation of posts traced by the man is
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x²/9 + y²/5 = 1
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x²/9 + y2 /25 = 1
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x²/5 + y²/9 = 1
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x²/25 + y²/9 = 1
Explanation
x²/25 + y²/9 = 1
Hint:
From the question, it is clear that the path traced by the man is an ellipse having its foci at two posts.
Let the equation of the ellipse be
x²/a² + y²/b² = 1
It is given that the sum of the distances of the man from the two flag posts is 10 m
This means that the sum of focal distances of a point on the ellipse is 10 m
⇒ PS + PS
1
= 10
⇒ 2a = 10
⇒ a = 5
Again, given that the distance between the flag posts is 8 meters
⇒ 2ae = 8
⇒ ae = 4
Now, b² = a² (1 - e²)
⇒ b² = a² - a² e²
⇒ b² = a² - (ae)²
⇒ b² = 5² - 4²
⇒ b² = 25 - 16
⇒ b² = 9
⇒ b = 3
Hence, the equation of the path is x²/5² + y²/3² = 1
⇒ x²/25 + y²/9 = 1
The center of the ellipse (x + y - 2)² /9 + (x - y)² /16 = 1 is
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(0, 0)
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(0, 1)
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(1, 0)
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(1, 1)
Explanation
(1, 1)
Hint:
The center of the given ellipse is the point of intersection of the lines
x + y - 2 = 0 and x - y = 0
After solving, we get
x = 1, y = 1
So, the center of the ellipse is (1, 1)
The parametric coordinate of any point of the parabola y² = 4ax is
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(-at², -2at)
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(-at², 2at)
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(a sin²t, -2a sin t)
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(a sin t, -2a sin t)
Explanation
(a sin²t, -2a sin t)
Hint:
The point (a sin²t, -2a sin t) satisfies the equation of the parabola y² = 4ax for all
values of t. So, the parametric coordinate of any point of the parabola y² = 4ax is
(a sin²t, -2a sin t)
The equation of parabola with vertex at origin the axis is along x-axis and passing through the point (2, 3) is
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y² = 9x
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y² = 9x/2
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y² = 2x
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y² = 2x/9
Explanation
y² = 9x/2
Hint:
A parabola with its axis along the x-axis and vertex(0, 0) and direction x = -a has the equation:
y² = 4ax .............. 1
Given, point (2,3) lies on the parabola,
⇒ 3² = 4a × 2
⇒ 9 = 4a × 2
⇒ 9/2 = 4a
From equation 1, we get
y² = (9/2)x
⇒ y² = 9x/2
This is the required equation of the parabola.
At what point of the parabola x² = 9y is the abscissa three times that of ordinate
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(1, 1)
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(3, 1)
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(-3, 1)
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(-3, -3)
Explanation
(3, 1)
Hint:
Given, parabola is x² = 9y
Let P(h, k) is the point on the parabola such that abscissa is 3 times the ordinate.
So, h = 3k ......... 1
Since P(h, k) lies on the parabola
So, h² = 9k ......... 2
From equation 1 and 2, we get
(3k)² = 9k
⇒ 9k² = 9k
⇒ 9k² - 9k = 0
⇒ 9k(k - 1) = 0
⇒ k = 0, 1
When k = 0, h = 0
So k = 1
Now, from equation 1,
h = 3 × 1 = 3
So, the point is (3, 1)
In an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is
0%
4/5
0%
1/√52
0%
3/5
0%
1/2
Explanation
3/5
Hint:
Given, distance between foci = 6
⇒ 2ae = 6
⇒ ae = 3
Again minor axis = 8
⇒ 2b = 8
⇒ b = 4
⇒ b² = 16
⇒ a² (1 - e²) = 16
⇒ a² - a² e² = 16
⇒ a² - (ae)² = 16
⇒ a² - 3² = 16
⇒ a² - 9 = 16
⇒ a² = 9 + 16
⇒ a² = 25
⇒ a = 5
Now, ae = 3
⇒ 5e = 3
⇒ e = 3/5
So, the eccentricity is 3/5
If the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is
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(x + 2)² + (y - 3)² = 3²
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(x - 2)² + (y + 3)² = 3²
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(x - 2)² + (y - 3)² = 3²
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(x + 2)² + (y + 3)² = 3²
Explanation
(x - 2)² + (y - 3)² = 3²
Hint:
Radius of the circle = √{(2 - 0)² + (3 - 0)² - 2²}
= √(4 + 9 - 4)
= √9
= 3
So, the equation of the circle = (x - 2)² + (y - 3)² = 3²
The equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is
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16x² - 9y² - 24xy - 144x + 8y + 224 = 0
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16x² + 9y² - 24xy - 144x + 8y - 224 = 0
0%
16x² + 9y² - 24xy - 144x - 8y + 224 = 0
0%
16x² + 9y² - 24xy - 144x + 8y + 224 = 0
Explanation
16x² + 9y² - 24xy - 144x + 8y + 224 = 0
Hint:
Given focus S(3, 0)
and equation of directrix is: 3x + 4y = 1
⇒ 3x + 4y - 1 = 0
Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix
Then, SP = PM
⇒ SP² = PM²
⇒ (x - 3)² + (y - 0)² = {(3x + 4y - 1) /{√(3² + 4²)}²
⇒ x² + 9 - 6x + y² = (9x² + 16y² + 1 + 24xy - 8y - 6x)/25
⇒ 25(x² + 9 - 6x + y²) = 9x² + 16y² + 1 + 24xy - 8y - 6x
⇒ 25x² + 225 - 150x + 25y² = 9x² + 16y² + 1 + 24xy - 8y - 6x
⇒ 25x² + 225 - 150x + 25y² - 9x² - 16y² - 1 - 24xy + 8y + 6x = 0
⇒ 16x² + 9y² - 24xy - 144x + 8y + 224 = 0
This is the required equation of parabola.
The parametric representation (2 + t², 2t + 1) represents
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a parabola
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a hyperbola
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an ellipse
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a circle
Explanation
a parabola
Hint:
Let x = 2 + t²
⇒ x - 2 = t² ........... 1
and y = 2t + 1
⇒ y - 1 = 2t
⇒ (y - 1)/2 = t
From equation 1, we get
x - 2 = {(y - 1)/2}²
⇒ x - 2 = (y - 1)²/4
⇒ (y - 1)² = 4(x - 2)
This represents the equation of a parabola.
The equation of a hyperbola with foci on the x-axis is
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x²/a² + y²/b² = 1
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x²/a² - y²/b² = 1
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x² + y² = (a² + b²)
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x² - y² = (a² + b²)
Explanation
x²/a² - y²/b² = 1
Hint:
The equation of a hyperbola with foci on the x-axis is defined as
x²/a² - y²/b² = 1
The equation of parabola with vertex (-2, 1) and focus (-2, 4) is
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10y = x² + 4x + 16
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12y = x² + 4x + 16
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12y = x² + 4x
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12y = x² + 4x + 8
Explanation
12y = x² + 4x + 16
Hint:
Given, parabola having vertex is (-2, 1) and focus is (-2, 4)
As the vertex and focus share the same abscissa i.e. -2,
parabola axis of symmetry as x = -2
⇒ x + 2 = 0
Hence, the equation of a parabola is of the type
(y - k) = a(x - h)² where (h, k) is vertex
Now, focus = (h, k + 1/4a)
Since, vertex is (-2, 1) and parabola passes through vertex
So, focus = (-2, 1 + 1/4a)
Now, 1 + 1/4a = 4
⇒ 1/4a = 4 -1
⇒ 1/4a = 3
⇒ 4a = 1/3
⇒ a = /1(3 × 4)
⇒ a = 1/12
Now, equation of parabola is
(y - 1) = (1/12) × (x + 2)²
⇒ 12(y - 1) = (x + 2)²
⇒ 12y - 12 = x² + 4x + 4
⇒ 12y = x² + 4x + 4 + 12
⇒ 12y = x² + 4x + 16
This is the required equation of parabola.
If a parabolic reflector is 20 cm in diameter and 5 cm deep then the focus of parabolic reflector is
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(0 0)
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(0, 5)
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(5, 0)
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(5, 5)
Explanation
(5, 0)
Hint:
given diameter of the parabola is 20 m.
The equation of parabola is y² = 4ax.
Since this parabola passes through the point A(5,10) then
10² = 4a×5
⇒ 20a = 100
⇒ a = 100/20
⇒ a = 5
So focus of parabola is (a, 0) = (5, 0)
The radius of the circle 4x² + 4y² - 8x + 12y - 25 = 0 is?
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√57/4
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√77/4
0%
√77/2
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√87/4
Explanation
√77/2
Hint:
Given, equation fo the of the circle is 4x² + 4y² - 8x + 12y - 25 = 0
⇒ x² + y² - 8x/4 + 12y/4 - 25/4 = 0
⇒ x² + y² - 2x + 3y - 25/4 = 0
Now, radius = √{(-2)² + (3)² - (-25/4)}
= √{4 + 9 + 25/4}
= √{13 + 25/4}
= √{(13×4 + 25)/4}
= √{(52 + 25)/4}
= √{77/4}
= √77/2
If (a, b) is the mid point of a chord passing through the vertex of the parabola y² = 4x, then
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a = 2b
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2a = b
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a² = 2b
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2a = b²
Explanation
2a = b²
Hint:
Let P(x, y) be the coordinate of the other end of the chord OP where O(0, 0)
Now, (x + 0)/2 = a
⇒ x = 2a
and (y + 0)/2 = b
⇒ y = 2b
Now, y² = 4x
⇒ (2b)² = 4 × 2a
⇒ 4b² = 8a
⇒ b² = 2a
A rod of length 12 CM moves with its and always touching the co-ordinate Axes. Then the equation of the locus of a point P on the road which is 3 cm from the end in contact with the x-axis is
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x²/81 + y²/9 = 1
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x²/9 + y²/81 = 1
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x²/169 + y²/9 = 1
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x²/9 + y²/169 = 1
Explanation
x²/81 + y²/9 = 1
Hint:
Given a rod of length 12 cm moves with its ends always touching the coordinate axes.
Again given a point P on the rod, which is 3 cm from the end in contact with the x-axis.
It is shown in the figure.
<img class="alignnone size-full wp-image-66272" src="https://www.learninsta.com/wp-content/uploads/2020/11/MCQ-Questions-for-Class-11-Maths-Chapter-11-Conic-Sections-with-Answers-3.jpg" alt="MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers 3" width="196" height="143" />
Here AP = 3 cm, AB = 12
Now BP = AB - AP
⇒ BP = 12 - 3
⇒ BP = 9 cm
Again from figure,
∠PAO = ∠BPO = θ (since PQ || OA and are corresponding angles)
Now in ΔBPO,
cosθ = QP/BP
⇒ cosθ = x/9 ............. 1
Again in ΔPAr,
sinθ = PR/PA
⇒ sinθ = y/3 ........ 2
Now square equation 1 and 2 and then add them, we get
cos² θ + sin² θ = x²/81 + y²/9
⇒ x²/81 + y²/9 = 1 (since cos² θ + sin² θ = 1 )
So, the equation of the locus of a point P is x²/81 + y²/9 = 1
The line lx + my + n = 0 will touches the parabola y² = 4ax if
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ln = am²
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ln = am
0%
ln = a² m²
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ln = a² m
Explanation
ln = am²
Hint:
Given, lx + my + n = 0
⇒ my = -lx - n
⇒ y = (-l/m)x + (-n/m)
This will touches the parabola y² = 4ax if
(-n/m) = a/(-l/m)
⇒ (-n/m) = (-am/l)
⇒ n/m = am/l
⇒ ln = am²
The center of the circle 4x² + 4y² - 8x + 12y - 25 = 0 is?
0%
(2,-3)
0%
(-2,3)
0%
(-4,6)
0%
(4,-6)
Explanation
(2,-3)
Hint:
Given, equation fo the of the circle is 4x² + 4y² - 8x + 12y - 25 = 0
⇒ x² + y² - 8x/4 + 12y/4 - 25/4 = 0
⇒ x² + y² - 2x + 3y - 25/4 = 0
Now, center = {-(-2), -3} = (2, -3)
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