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Test: Parabola - JEE MCQ


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

Test: Parabola for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Parabola questions and answers have been prepared according to the JEE exam syllabus.The Test: Parabola MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Parabola below.
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Test: Parabola - Question 1

The equation of parabola whose focus is (– 3, 0) and directrix x + 5 = 0 is:

Detailed Solution for Test: Parabola - Question 1

According to definition of parabola , is is the locks of the points in that planes that are equidistant from both focus and directrix.

Given, focus : (-3,0)
directrix : x + 5 = 0
Let (x ,y) is the point on the parabola .
∴ distance of point from focus = distance of point from directrix
⇒ √{(x + 3)2 + y2} = |x + 5|/√(12 + 02)
⇒ √{(x + 3)2 + y2 } = |x + 5|
squaring both sides,
(x + 3)2 + y2 = (x + 5)2
⇒y2 = (x + 5)2 - (x + 3)2
⇒y2= (x + 5 - x - 3)(x + 5 + x + 3)
⇒y2 = 2(2x + 8) = 4(x + 4)

Hence, equation of parabola is y2 = 4(x + 4)

Test: Parabola - Question 2

A parabola whose axis is along the y-axis, vertex is (0,0) and point from the first and second quadrants lie on it, has the equation of the type

Detailed Solution for Test: Parabola - Question 2

As the quadrants lies in first and second quadrant
y = -a    Focus(0,a)
x2 = 4ay
 

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Test: Parabola - Question 3

The equation y2 + 3 = 2(2x+y) represents a parabola with the vertex at

Detailed Solution for Test: Parabola - Question 3

y2+3=2(2x+y) represents parabola.
y2+3=4x+2y
y2−2y+3=4x
y2−2y+1+3=4x+1
(y−1)2=4x−2
(y−1)2=4(x−1/2)
So, the vertex of parabola=(1/2,1) and axis is parallel to x axis.
a=1
Focus=(1/2+1,1)
=(3/2,1)

Test: Parabola - Question 4

The equation of the parabola with vertex at (0, 0) and focus at (0, – 2) is:

Detailed Solution for Test: Parabola - Question 4

Given the vertex of the parabola is (0,0) and focus is at (0,-2).
This gives the axis of the parabola is the positive y− axis.
Then the equation of the parabola will be x^2 = 4ay where a = -2.
So the equation of the parabola is x2 = -8y.

Test: Parabola - Question 5

The …… of a conic is the chord passing through the focus and perpendicular to the axis.

Detailed Solution for Test: Parabola - Question 5

The chord of a parabola through the focus and perpendicular to the axis is called the latus rectum.

Test: Parabola - Question 6

The equation of the parabola with vertex at (1, 2) and focus at (2, 0) is:

Detailed Solution for Test: Parabola - Question 6



Test: Parabola - Question 7

The equation of the parabola with focus at (a,0) where a>0 and directrix x = -a is

Detailed Solution for Test: Parabola - Question 7

When a is positive, the parabola is opening to the right and since directrix is x=-a, it gives the answer.

Test: Parabola - Question 8

If the focus of a parabola is (-2,1) and the directrix has the equation x + y = 3 then the vertex is

Detailed Solution for Test: Parabola - Question 8

x+y=3 m=−1
(−2,1)m = 1
y−1 = 1(x+2)
y−1 = x+2
x−y+3 = 0
x−y+3 = 0
x+y−3 = 0
2x = 0
x = 0; y = 3
Vertex=
Midpoint of focus and directrix = (0−2)/2,(3+1)/2
​=(−1,2)

Test: Parabola - Question 9

Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given by

Detailed Solution for Test: Parabola - Question 9

f(0,1),d(x+2=0)
Distance of any point on parabola and focus is equal to distance of point and directrix.
fP=(h−0)2+(k−1)2= (h2+k2+1−2k)1/2
Distance of point (h,k) and line x+2=0
Using point line distance formula.
dP=h+2
[h2+k2+1−2k]1/2=h+2
h2+k2+1−2k = h2+4+4h
k2−2k+1−4−4h=0
replacing h→x,k→y  y2−2y+1−4−4x=0
(y−1)2=4(x+1)     …(1)
Let Y=y−1,X=x+1 then (1) becomes 
Y^2=4aX2
Here a=1 any point on this parabola will be of the form (at2,2at)=(t2,2at)
⇒X=t2 ⇒x+1=t2
⇒x=t2−1
⇒Y2=2t
⇒y−1 = 2t ⇒ y = 2t+1
∴ Any point on the parabola (y−1)2=4(x+1) is 
= (t2−1,2t+1)

Test: Parabola - Question 10

The equation 2x2 – 3xy + 5y2 + 6x – 3y + 5 = 0 represents.

Detailed Solution for Test: Parabola - Question 10

Comparing the equation with the standard form ax2+2hxy+by2+2gx+2fy+c=0
a=2,h=−3/2,b=5,g=3,f=−3/2,c=5
Δ=abc+2fgh−af2−bg2−ch2
=(2)(5)(5)+2(−3/2)(3)(−3/2)−(2)(−3/2)2−(5)(3)2−(5)(−3/2)2
=50+27/2−9/2−45−225/4
=−169/4 is not equal to 0
Descriminant =h2−ab
=(−3/2)2−(2)(5)
= 9/4−10
= −31/4<0
So, the curve represents either a circle or an ellipse
a is not equal to b and  
Δ/a+b = −(169/4)/2+5
=−169/28<0
So, the curve represents a ellipse.

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