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Test: Parabola- 2 - JEE MCQ


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20 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Parabola- 2

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Test: Parabola- 2 - Question 1

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

Test: Parabola- 2 - Question 2

Directrix of a parabola is x + y = 2. If it's focus is origin, then latus rectum of the parabola is equal to

Detailed Solution for Test: Parabola- 2 - Question 2

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 

Test: Parabola- 2 - Question 3

Which one of the following equations represents parametrically, parabolic profile ?

Detailed Solution for Test: Parabola- 2 - Question 3

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.

Test: Parabola- 2 - Question 4

If (t2, 2t) is one end of a focal chord of the parabola y2 = 4x then the length of the focal chord will be

Detailed Solution for Test: Parabola- 2 - Question 4

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

Test: Parabola- 2 - Question 5

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

Detailed Solution for Test: Parabola- 2 - Question 5

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

Test: Parabola- 2 - Question 6

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

Detailed Solution for Test: Parabola- 2 - Question 6

For intersection of both the curve we must have,

Therefore, the point of intersection is (2,2)(2,2)
Hence, option 'B' is correct.

Test: Parabola- 2 - Question 7

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)

Detailed Solution for Test: Parabola- 2 - Question 7

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)/(at2/4 - 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

Test: Parabola- 2 - Question 8

The tangents to the parabola x = y2 + c from origin are perpendicular then c is equal to

Detailed Solution for Test: Parabola- 2 - Question 8

Test: Parabola- 2 - Question 9

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

Detailed Solution for Test: Parabola- 2 - Question 9

Let the point be (h, k)

Now equation of tangent to the parabola y= 4ax whose slope is m is

as it passes through (h, k)]

Test: Parabola- 2 - Question 10

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

Detailed Solution for Test: Parabola- 2 - Question 10


Test: Parabola- 2 - Question 11

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

Test: Parabola- 2 - Question 12

The equation of the tangent at the vertex of the parabola x2 + 4x + 2y = 0 is

Detailed Solution for Test: Parabola- 2 - Question 12

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

Test: Parabola- 2 - Question 13

Locus of the point of intersection of the perpendicular tangents of the curve y2 + 4y – 6x – 2 = 0 is

Detailed Solution for Test: Parabola- 2 - Question 13

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)
Y= 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

Test: Parabola- 2 - Question 14

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

Detailed Solution for Test: Parabola- 2 - Question 14

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)

Test: Parabola- 2 - Question 15

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

Detailed Solution for Test: Parabola- 2 - Question 15

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)

Test: Parabola- 2 - Question 16

If (3t12-6t1) represents the feet of the normals to the parabola y2 = 12x from (1, 2), then Σ1/t1 is

Test: Parabola- 2 - Question 17

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

Test: Parabola- 2 - Question 18

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

Test: Parabola- 2 - Question 19

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)

Detailed Solution for Test: Parabola- 2 - Question 19

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

Test: Parabola- 2 - Question 20

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

Detailed Solution for Test: Parabola- 2 - Question 20

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

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