Conic Sections Test

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Conic Sections Test - 4

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Weekly Quiz Competition

Question 1
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If the line x + y –1 = 0 touches the parabola y² = kx , then the value of k is

A
4

B
–4

C
2

D
–2

Solution

If y = -x + 1 is tangent to y² = kx, then
c = a / m
c = (k / 4) / (-1)
1 = (k / 4) / (-1)
k = -4
Question 2
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Directrix of a parabola is x + y = 2. If it 's focus is origin, then latus rectum of the parabola is equal to

A
√2 units

B
2 units

C
4 √2 units

D
4 units

Solution

Given directrix of parabola ⇒x+y=2
and force is origin vertex is A(0,0)
We know that perpendicular distance from vertex of the parabola to directrix is equal to 'a ' where 4a is the Latus Rectum of the
Parabola
Question 3
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Which one of the following equations represents parametrically, parabolic profile ?

A
x = 3 cost ; y = 4 sint

B
x² – 2 = – cost ; y = 4 cos² t/2

C
√x = tan t ; √y = sec t

D

Solution

x² −2 = −2cost
⇒x² = 2 −2cost
⇒x² = 2(1 −cost)
⇒x² = 2(1 −(1 −2sin² t/2))
⇒x² = 4sin² t/2
We have y = 4cos² t/2
⇒cos² t/2= y/4
We know the identity, sin² t/2 + cos² t/2 = 1
⇒x² /4 + y/4 = 1
⇒x² = 4 −y represents a parabolic profile.
Question 4
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If (t² , 2t) is one end of a focal chord of the parabola y² = 4x then the length of the focal chord will be

A

B

C

D
None

Solution

Given (t² , 2t) be one end of focal chord then other end be (1/t² , −2t )

Length of focal chord = [(t² - 1/t² )² + (2t + t/2)² ]^½

= ( t + 1/t)²
Question 5
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From the focus of the parabola y² = 8x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is

A
(x – 2)² + y² = 3

B
(x – 2)² + y² = 9

C
(x + 2)² + y² = 9

D
None

Solution

Focus of parabola y² = 8x is (2,0). Equation of circle with centre (2,0) is (x −2)² + y² = r²
Let AB is common chord and Q is mid point i.e. (1,0)
AQ² = y² = 8x
= 8 ×1 = 8
∴r² = AQ² + QS²
= 8 + 1 = 9
So required circle is (x −2)² + y² = 9
Question 6
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The point of intersection of the curves whose parametric equations are x = t² + 1, y = 2t and x = 2s, y = 2/s is given by

A
(1, –3)

B
(2, 2)

C
(–2, 4)

D
(1, 2)

Solution

For intersection of both the curve we must have,

Therefore, the point of intersection is (2,2)(2,2)
Hence, option 'B 'is correct.
Question 7
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PN is an ordinate of the parabola y² = 4ax. A straight line is drawn parallel to the axis to bisect NP and meets the curve in Q. NQ meets the tangent at the vertex in a point T such that AT = kNP, then the value of k is (where A is the vertex)

A
3/2

B
2/3

C
1

D
none

Solution

The equation of parabola be y² =4ax
let the point P be (at² ,2at)
PN is ordinate ⇒N(at² ,0)
Equation of straight line bisecting NP is
y=at
substituting y in equation of parabola
a² t² = 4ax
⇒x = 4at²
So the coordinates of Q are (4at² ,at)
Equation of NQ is y −0 = (at −0)/(at^2/4 - at² )(x −at² )
y= −4/3t(x −at² )
Put x=0
y = −4/3t(0 −at^2)
⇒y=4at/3
⇒AT = 4at/3
NP = 2at
AT/NP = (4at/3)/2at
= ⅔
AT = 2/3NP
Question 8
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The tangents to the parabola x = y² + c from origin are perpendicular then c is equal to

A
1/2

B
1

C
2

D
1/4

Solution
Question 9
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The locus of a point such that two tangents drawn from it to the parabola y² = 4ax are such that the slope of one is double the other is

A
y² = 9/2 ax

B
y² = 9/4 ax

C
y² = 9ax

D
x² = 4ay

Solution

Let the point be (h, k)
Now equation of tangent to the parabola y² = 4ax whose slope is m is

as it passes through (h, k)]
Question 10
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T is a point on the tangent to a parabola y² = 4ax at its point P. TL and TN are the perpendiculars on the focal radius SP and the directrix of the parabola respectively. Then

A
SL = 2 (TN)

B
3 (SL) = 2 (TN)

C
SL = TN

D
2 (SL) = 3 (TN)

Solution
Question 11
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The equation of the circle drawn with the focus of the parabola (x – 1)² – 8y = 0 as its centre and touching the parabola at its vertex is

A
x² + y² – 4y = 0

B
x² + y² – 4y + 1 = 0

C
x² + y² – 2x – 4y = 0

D
x² + y² – 2x – 4y + 1 = 0
Question 12
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The equation of the tangent at the vertex of the parabola x² + 4x + 2y = 0 is

A
x = –2

B
x = 2

C
y = 2

D
y = –2.

Solution

The Equation of tangent at vertex to parabola x² +4x+2y=0 is :
x² +4x+2y+4 −4=0
(x+2)² = −2(y −2)
x² =−2y
Equation of tangent at vertex is y=0
y −2=0
y=2
Question 13
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Locus of the point of intersection of the perpendicular tangents of the curve y² + 4y – 6x – 2 = 0 is

A
2x – 1 = 0

B
2x + 3 = 0

C
2x + 5 = 0

D
2y + 3 = 0

Solution

Given parabola is, y² +4y −6x −2=0
⇒y² +4y+4=6x+6=6(x+1)
⇒(y+2)² = 6(x+1)
shifting origin to (−1,−2)
Y² = 4aX where a = 3/2
We know locus of point of intersection of perpendicular tangent is directrix of the parabola itself
Hence required locus is X=−a ⇒x+1=−3/2
⇒2x+5=0
Question 14
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Tangents are drawn from the points on the line x – y + 3 = 0 to parabola y² = 8x. Then the variable chords of contact pass through a fixed point whose coordinates are

A
(3, 2)

B
(2, 4)

C
(3, 4)

D
(4, 1)

Solution

Let (k,k+3) be the point on the line x −y+3=0
Equation of chord of contact is S₁ =0
⇒yy1=4(x+x1)
⇒y(k+3)=4(x+k)
⇒4x −3y −k(y −4)=0
Therefore, straight line passes through fixed point (3,4)
Question 15
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The line 4x – 7y + 10 = 0 intersects the parabola, y² = 4x at the points A &B. The co-ordinates of the point of intersection of the tangents drawn at the points A &B are

A

B

C

D

Solution

ky=[4(x+h)]/2
=>2ky=2(x+h)
2ky=4x+4h =>4x −2ky+4h=0
4x −7y+10=0
4h=10 =>h=5/2
2k=7 =>k=7/2
point of intersection of tan ⁡gent at p and q is (5/2,7/2)
Question 16
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If (3t₁² -6t₁ ) represents the feet of the normals to the parabola y² = 12x from (1, 2), then Σ1/t₁ is

A
–5/2

B
3/2

C
6

D
–3
Question 17
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TP &TQ are tangents to the parabola, y² = 4ax at P &Q. If the chord PQ passes through the fixed point (–a, b) then the locus of T is

A
ay = 2b (x – b)

B
bx = 2a (y – a)

C
by = 2a (x – a)

D
ax = 2b (y – b)
Question 18
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If the tangent at the point P (x₁ , y₁ ) to the parabola y² = 4ax meets the parabola y² = 4a (x + b) at Q &R, then the mid point of QR is

A
(x₁ + b, y₁ + b)

B
(x₁ - b, y₁ - b)

C
(x₁ , y₁ )

D
(x₁ + b, y₁ )
Question 19
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Let PSQ be the focal chord of the parabola, y² = 8x. If the length of SP = 6 then, l(SQ) is equal to(where S is the focus)

A
3

B
4

C
6

D
None

Solution

Since the semi latus rectum of a parabola is the harmonic mean between the segment of any focal chord of a parabola, therefore,SP,4,SQ are in H.P.
⇒4=2(SP.SQ)/(SP+SQ)
⇒4=2*6.SQ/(6+SQ)
⇒SQ=3
Question 20
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Two parabolas y² = 4a(x – l₁ ) and x² = 4a(y – l₂ ) always touch one another, the quantities l₁ and l₂ are both variable. Locus of their point of contact has the equation

A
xy = a²

B
xy = 2a²

C
xy = 4a²

D
None

Solution

Let P(x₁ , y₁ ) be point of contact of two parabola. tangents at P of the two parabolas are