Parabola Test

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Parabola Test - 7

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Question 1
4 / -1

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 2
4 / -1

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
x = ; y = sin t/2 + cos 1/2

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 3
4 / -1

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 4
4 / -1

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 5
4 / -1

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 6
4 / -1

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
a²t² = 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 7
4 / -1

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 8
4 / -1

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 9
4 / -1

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 10
4 / -1

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 11
4 / -1

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 12
4 / -1

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 13
4 / -1

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 14
4 / -1

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 15
4 / -1

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 16
4 / -1

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