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All questions of Coordinate Geometry- Parabola for A Level Exam
The equation of the parabola with vertex at (0, 0) and focus at (0, – 2) is:- a)y2 = – 2x
- b)x2 = – 8y
- c)y2 = – 8x
- d)x2 = – 4y
Correct answer is option 'B'. Can you explain this answer?
The equation of the parabola with vertex at (0, 0) and focus at (0, – 2) is:
a)
y2 = – 2x
b)
x2 = – 8y
c)
y2 = – 8x
d)
x2 = – 4y
|
Praveen Kumar answered |
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.
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.
Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given by- a)(t2 + 1, 2t – 1)
- b)(t2, 2t)
- c)(t2 + 1, 2t + 1)
- d)(t2 – 1, 2t + 1)
Correct answer is option 'D'. Can you explain this answer?
Any point on the parabola whose focus is (0,1) and the directrix is x + 2 = 0 is given by
a)
(t2 + 1, 2t – 1)
b)
(t2, 2t)
c)
(t2 + 1, 2t + 1)
d)
(t2 – 1, 2t + 1)
|
Pooja Shah answered |
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)
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)
A variable chord PQ of the parabola y = 4x2 substends a right angle at the vertex. Then the locus of points of intersection of the tangents at P and Q is- a)4y + 1= 16x2
- b)y + 4 = 0
- c)4y + 4 = 4x2
- d)4y + 1 = 0
Correct answer is option 'D'. Can you explain this answer?
A variable chord PQ of the parabola y = 4x2 substends a right angle at the vertex. Then the locus of points of intersection of the tangents at P and Q is
a)
4y + 1= 16x2
b)
y + 4 = 0
c)
4y + 4 = 4x2
d)
4y + 1 = 0
|
Krishna Iyer answered |
Slope of OP x slope of OQ = -1
⇒ 4t1.4t2 = -1
Eq of tangent at
Eq of tangent at
Let (x1 , y1) is the point of intersection
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)
Correct answer is option 'C'. Can you explain this answer?
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)
|
Yash Patel answered |
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 tangent at p and q is (5/2,7/2)
=> 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 tangent at p and q is (5/2,7/2)
The equation 
represents a parabola with the vertex at- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The equation represents a parabola with the vertex at
represents a parabola with the vertex ata)
b)
c)
d)
|
Gaurav Kumar answered |
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)
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)
The straight line joining any point P on the parabola y2 = 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R is- a) x2 + 2y2 – ax = 0
- b)2x2 + y2 – 2ax = 0
- c)2x2 + 2y2 – ay = 0
- d)2x2 + y2 – 2ay = 0
Correct answer is option 'B'. Can you explain this answer?
The straight line joining any point P on the parabola y2 = 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R is
a)
x2 + 2y2 – ax = 0
b)
2x2 + y2 – 2ax = 0
c)
2x2 + 2y2 – ay = 0
d)
2x2 + y2 – 2ay = 0
|
|
Krishna Iyer answered |
Given the equation of parabola is
y2 = 4ax
Let P(at2, 2at) be any point on the parabola.
Equation of tangent at point P is ty = x + at2 where slope of the tangent is 1/t.
Equation of line perpendicular to the tangent passes through (a,0) is given as
∴ y−0 = −t(x−a)
or y = t(a−x) .....(i)
Equation of OP is given by
y−0 = 2/t(x−0) = 0
⇒ y = 2x/t .....(ii)
Eliminating 't' from equations (i) and (ii), we get
y2 = 2x(a−x)
or 2x2 + y2 − 2ax = 0
y2 = 4ax
Let P(at2, 2at) be any point on the parabola.
Equation of tangent at point P is ty = x + at2 where slope of the tangent is 1/t.
Equation of line perpendicular to the tangent passes through (a,0) is given as
∴ y−0 = −t(x−a)
or y = t(a−x) .....(i)
Equation of OP is given by
y−0 = 2/t(x−0) = 0
⇒ y = 2x/t .....(ii)
Eliminating 't' from equations (i) and (ii), we get
y2 = 2x(a−x)
or 2x2 + y2 − 2ax = 0
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
Correct answer is option 'B'. Can you explain this answer?
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
|
|
Geetika Shah answered |
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
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
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- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
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
a)
b)
c)
d)
|
Rohini Verma answered |
As the quadrants lies in first and second quadrant
y = -a Focus(0,a)
x2 = 4ay
y = -a Focus(0,a)
x2 = 4ay
If the tangents and normals at the extremities of a focal chord of a parabola intersect at (x1, y1) and(x2, y2) respectively, then- a) x1 = x2
- b) x1 = y2
- c)y1 = y2
- d)x2 = y1
Correct answer is option 'C'. Can you explain this answer?
If the tangents and normals at the extremities of a focal chord of a parabola intersect at (x1, y1) and
(x2, y2) respectively, then
a)
x1 = x2
b)
x1 = y2
c)
y1 = y2
d)
x2 = y1
|
Neha Joshi answered |
If the parabola is Y2= 4ax
take the focal chord which is easy for calculation e.x. LR (latus rectum)
then coordinates of extremities would be (a,2a) and (a,-2a)
equation of tangent of parabola at (a,2a) :
T=0 : 2ay = 2a(x+a)
y = x+a................................. (1)
equation of tangent of parabola at (a,-2a) :
T=0 -2ay = 2a (x+a)
y = -x-a .........................(2)
point of intersection of both tangents is (X1, Y1)
after solving eq1 and eq2 X1 = -a and Y1 = 0
so ( -a, 0)
eqn of normal of parabola at (a, 2a)
y = -x +3a ...............................(3)
eqn of normal of parabola at (a, -2a)
y = x -3a..............................(4)
so point of intersection of normal's : (X2, Y2)
after solving eq3 and eq4 X2= 3a and Y2 = 0
so we conclude... for y2= 4ax
Y1= Y2
take the focal chord which is easy for calculation e.x. LR (latus rectum)
then coordinates of extremities would be (a,2a) and (a,-2a)
equation of tangent of parabola at (a,2a) :
T=0 : 2ay = 2a(x+a)
y = x+a................................. (1)
equation of tangent of parabola at (a,-2a) :
T=0 -2ay = 2a (x+a)
y = -x-a .........................(2)
point of intersection of both tangents is (X1, Y1)
after solving eq1 and eq2 X1 = -a and Y1 = 0
so ( -a, 0)
eqn of normal of parabola at (a, 2a)
y = -x +3a ...............................(3)
eqn of normal of parabola at (a, -2a)
y = x -3a..............................(4)
so point of intersection of normal's : (X2, Y2)
after solving eq3 and eq4 X2= 3a and Y2 = 0
so we conclude... for y2= 4ax
Y1= Y2
Which one of the following equations represents parametrically, parabolic profile ?- a)x = 3 cost ; y = 4 sint
- b) x2 – 2 = – cost ; y = 4 cos2
- c)
= tan t ; 
= sec t - d) x =
; y = sin
+ cos
Correct answer is option 'B'. Can you explain this answer?
Which one of the following equations represents parametrically, parabolic profile ?
a)
x = 3 cost ; y = 4 sint
b)
x2 – 2 = – cost ; y = 4 cos2
c)
= tan t ; = sec td)
x = ; y = sin + cos
; y = sin + cos
|
Hansa Sharma answered |
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.
⇒ 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.
If the focus of a parabola is (-2,1) and the directrix has the equation x + y = 3 then the vertex is- a)(2,-1)
- b)(-1,2)
- c)(0,3)
- d)(-1,1/2)
Correct answer is option 'B'. Can you explain this answer?
If the focus of a parabola is (-2,1) and the directrix has the equation x + y = 3 then the vertex is
a)
(2,-1)
b)
(-1,2)
c)
(0,3)
d)
(-1,1/2)
|
|
Riya Banerjee answered |
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)
(−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)
The …… of a conic is the chord passing through the focus and perpendicular to the axis.- a)Latus-rectum
- b)Directrix
- c)Line of axis
- d)Y-axis
Correct answer is option 'A'. Can you explain this answer?
The …… of a conic is the chord passing through the focus and perpendicular to the axis.
a)
Latus-rectum
b)
Directrix
c)
Line of axis
d)
Y-axis
|
Jitendra Meena answered |
Answer is A. As latus rectum is a line that passes through the focus and also is perpendicular to the axis
The parametric coordinates of any point on the parabola y2 = 4ax can be- a)(at2, 2at)
- b)(at2, –2at)
- c)(asin2t, 2asint)
- d)(asint, 2acost)
Correct answer is option 'A'. Can you explain this answer?
The parametric coordinates of any point on the parabola y2 = 4ax can be
a)
(at2, 2at)
b)
(at2, –2at)
c)
(asin2t, 2asint)
d)
(asint, 2acost)
|
Lalit Yadav answered |
Parametric coordinates of the parabola y2 = 4ax are (at2, 2at).
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 =
ax - b) y2 =
ax - c) y2 = 9ax
- d)x2 = 4ay
Correct answer is option 'A'. Can you explain this answer?
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 = ax
axb)
y2 = ax
axc)
y2 = 9ax
d)
x2 = 4ay
|
Knowledge Hub answered |
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)]
as it passes through (h, k)]
The equation 2x2 – 3xy + 5y2 + 6x – 3y + 5 = 0 represents.- a)A parabola
- b)An ellipse
- c)A hyperbola
- d)A pair of straight lines
Correct answer is option 'B'. Can you explain this answer?
The equation 2x2 – 3xy + 5y2 + 6x – 3y + 5 = 0 represents.
a)
A parabola
b)
An ellipse
c)
A hyperbola
d)
A pair of straight lines
|
|
Pooja Shah answered |
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.
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.
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)
Correct answer is option 'C'. Can you explain this answer?
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)
|
Arnav Kulkarni answered |
D
The straight line x + y = k + 1 touches the parabola y = x(1 – x) if- a)k = -1
- b)k = 0
- c)k = 1
- d)k takes any real value
Correct answer is option 'B'. Can you explain this answer?
The straight line x + y = k + 1 touches the parabola y = x(1 – x) if
a)
k = -1
b)
k = 0
c)
k = 1
d)
k takes any real value
|
Nishanth Verma answered |
Method to Solve :
x+y-1=0. or. y=1-x (1)
y^2. = kx .(2)
On putting y=1-x from eq.(1)
(1-x)^2=k.x
1–2x+x^2 =kx
x^2 - (k+2)x+1 = 0
The given line is tangent to the parabola which touches the parabola one and only one point. means both the roots are same.
D or. b^ - 4 a.c. =0
{-(k+2)}^2. - 4.1.1. =0
(k+2)^2. = 4
(k+2) = +/-(2)
k= -2 +/-2
k= -2+2 or. -2–2.
k =0. , - 4. Answer.
y^2. = kx .(2)
On putting y=1-x from eq.(1)
(1-x)^2=k.x
1–2x+x^2 =kx
x^2 - (k+2)x+1 = 0
The given line is tangent to the parabola which touches the parabola one and only one point. means both the roots are same.
D or. b^ - 4 a.c. =0
{-(k+2)}^2. - 4.1.1. =0
(k+2)^2. = 4
(k+2) = +/-(2)
k= -2 +/-2
k= -2+2 or. -2–2.
k =0. , - 4. Answer.
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)2y + 3 = 0
- d)2x + 5 = 0
Correct answer is option 'C'. Can you explain this answer?
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)
2y + 3 = 0
d)
2x + 5 = 0
|
|
Geetika Shah answered |
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
⇒ 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
If the line 2x – y + λ = 0 is a diameter of the circle x2+y2+6x−6y+5 = 0 then λ =- a)6
- b)9
- c)3
- d)12
Correct answer is option 'B'. Can you explain this answer?
If the line 2x – y + λ = 0 is a diameter of the circle x2+y2+6x−6y+5 = 0 then λ =
a)
6
b)
9
c)
3
d)
12
|
|
Rajesh Gupta answered |
x2 + y2 + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)2 + (3)2 - 5}
= √{9 + 9 - 5}
= √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
=> -3*2 - 3 + λ = 0
=> -6 - 3 + λ = 0
=> -9 + λ = 0
=> λ = 9
Center O = (-3, 3)
radius r = √{(-3)2 + (3)2 - 5}
= √{9 + 9 - 5}
= √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
=> -3*2 - 3 + λ = 0
=> -6 - 3 + λ = 0
=> -9 + λ = 0
=> λ = 9
Focus and vertex of the parabola that touches x-axis at (1, 0) and x = y at (1, 1) are (h, k) and (p, q) then the value of 25(p q h k)
- a) 35
- b) 37
- c) 33
- d) 39
Correct answer is option 'B'. Can you explain this answer?
Focus and vertex of the parabola that touches x-axis at (1, 0) and x = y at (1, 1) are (h, k) and (p, q) then the value of 25(p q h k)
a)
35b)
37c)
33d)
39|
|
Poonam Reddy answered |
The x-axis touches at A(1, 0) and x = y touches at B(1, 1). Hence the equation to the curve through these points is given by y(y – x) + k(x – 1)2 = 0. For this to represent a parabola, 4k = 1. The equation is x2 – 4xy + 4y2 – 2x + 1 = 0. Vertex 
focus 
focus 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)
Correct answer is option 'B'. Can you explain this answer?
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)
|
|
Bhargavi Joshi answered |
For intersection of both the curve we must have,
Therefore, the point of intersection is (2,2)(2,2)
Hence, option 'B' is correct.
Therefore, the point of intersection is (2,2)(2,2)
Hence, option 'B' is correct.
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
Correct answer is option 'B'. Can you explain this answer?
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
|
|
Naina Sharma answered |
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
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
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
Correct answer is option 'C'. Can you explain this answer?
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
|
|
Aaradhya Solankir answered |
Ritu Singh ,
There is not answer in the given options. What Saroj Bala said is wrong because there it's considered the vertex as origin , but in the question the focus is given as origin...
There is not answer in the given options. What Saroj Bala said is wrong because there it's considered the vertex as origin , but in the question the focus is given as origin...
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
Correct answer is option 'C'. Can you explain this answer?
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
|
|
Gaurav Kumar answered |
Let P(x1 , y1) be point of contact of two parabola. tangents at P of the two parabolas are
In a knockout tournament 16 equally skilled players namely P1, P2, -------- P16 are participating. In each round players are divided in pairs at random and winner from each pair moves in the next round. If P2 reaches the semifinal, then the probability that P1 will win the tournament is.- a)3/64
- b)1/16
- c)1/20
- d)1/15
Correct answer is option 'C'. Can you explain this answer?
In a knockout tournament 16 equally skilled players namely P1, P2, -------- P16 are participating. In each round players are divided in pairs at random and winner from each pair moves in the next round. If P2 reaches the semifinal, then the probability that P1 will win the tournament is.
a)
3/64
b)
1/16
c)
1/20
d)
1/15
|
Anjali Sharma answered |
Let E1 = P1 win the tournament, E2 = P2 reaches the semifinal since all players are equally skilled and there are 4 persons in the semifinal 
both are in semifinal and P1 wins in semifinal and final
both are in semifinal and P1 wins in semifinal and final The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is- a)1
- b)0
- c)3
- d)2
Correct answer is option 'B'. Can you explain this answer?
The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is
a)
1
b)
0
c)
3
d)
2
|
Mohit Rajpoot answered |
Distance of 'c' units from (2,3)
Let the no: of points be (x,0)
By distance formula
{(2−x)2+(3−0)2}=c
4−4x+x2+9=c
⇒x2−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.
Let the no: of points be (x,0)
By distance formula
{(2−x)2+(3−0)2}=c
4−4x+x2+9=c
⇒x2−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.
The length of the chord of the parabola y2 = x which is bisected at the point (2, 1) is less than- a)5√2
- b)4√5
- c)
- d)2√5
Correct answer is option 'A,B,C,D'. Can you explain this answer?
The length of the chord of the parabola y2 = x which is bisected at the point (2, 1) is less than
a)
5√2
b)
4√5
c)
d)
2√5
|
Veda Institute answered |
Equation of chord to the given parabola with given mid point (2,1) is given by,
T = S1
T = S1
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.
Correct answer is option 'C'. Can you explain this answer?
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.
|
Akshay Chauhan answered |
-2
To find the equation of the tangent at the vertex of the parabola, we need to first find the coordinates of the vertex. We can do this by completing the square:
x^2 + 4x + 2y = 0
x^2 + 4x + 4 + 2y = 4
(x + 2)^2 = -2y + 4
So the vertex is at (-2, 2).
Next, we need to find the slope of the tangent at the vertex. We can do this by taking the derivative of the equation of the parabola:
2x + 4 + 0 = 0
2x = -4
x = -2
So the slope of the tangent at the vertex is 2(-2) + 4 = 0.
Finally, we can use the point-slope form of a line to find the equation of the tangent:
y - 2 = 0(x + 2)
y - 2 = 0
y = 2
So the equation of the tangent at the vertex is y = 2.
To find the equation of the tangent at the vertex of the parabola, we need to first find the coordinates of the vertex. We can do this by completing the square:
x^2 + 4x + 2y = 0
x^2 + 4x + 4 + 2y = 4
(x + 2)^2 = -2y + 4
So the vertex is at (-2, 2).
Next, we need to find the slope of the tangent at the vertex. We can do this by taking the derivative of the equation of the parabola:
2x + 4 + 0 = 0
2x = -4
x = -2
So the slope of the tangent at the vertex is 2(-2) + 4 = 0.
Finally, we can use the point-slope form of a line to find the equation of the tangent:
y - 2 = 0(x + 2)
y - 2 = 0
y = 2
So the equation of the tangent at the vertex is y = 2.
A parabola has focus at (0, 0) and passes through the points (4, 3) and (–4, –3). The number of lattice points (x, y) on the parabola such that |4x + 3y| < 1000 is- a)20
- b)100
- c)40
- d)80
Correct answer is option 'D'. Can you explain this answer?
A parabola has focus at (0, 0) and passes through the points (4, 3) and (–4, –3). The number of lattice points (x, y) on the parabola such that |4x + 3y| < 1000 is
a)
20
b)
100
c)
40
d)
80
|
Devansh Joshi answered |
The expression you'll find is 4x+3y=1250n+50p−25, n being a integer and p being a positive integer that is less than 125050
Then, you can build a simple excel file to exhaust the answers from 1250n+25
Then, you can build a simple excel file to exhaust the answers from 1250n+25
to 1250n+1225. It turns out that only when n=0 or 1 are the criteria satisfied, and in each case there are 20 of the pairs.
So B.40
The equation of a straight line passing through the point (3, 6) and cutting the curve y = √x orthogonally is- a)4x + y – 18 = 0
- b)x + y – 9 = 0
- c)4x – y – 6 = 0
- d)None
Correct answer is option 'A'. Can you explain this answer?
The equation of a straight line passing through the point (3, 6) and cutting the curve y = √x orthogonally is
a)
4x + y – 18 = 0
b)
x + y – 9 = 0
c)
4x – y – 6 = 0
d)
None
|
Sakshi Sen answered |
The curve y=(x)1/2 is the part of curve y2=x
Equation of normal at P(t2/4, t/2) is
y+tx=t/2+t3/4..... (1)
The equation will pass through (3,6)
6+3t=t/2+t3/4
t3−10t−24=0
Solving, we get t=4
So equation of line which is orthogonal and passes through (3,6) is
y+4x=18
4x+y-18 = 0
Equation of normal at P(t2/4, t/2) is
y+tx=t/2+t3/4..... (1)
The equation will pass through (3,6)
6+3t=t/2+t3/4
t3−10t−24=0
Solving, we get t=4
So equation of line which is orthogonal and passes through (3,6) is
y+4x=18
4x+y-18 = 0
The locus of the centre of a circle, which touches externally the circle x2 + y2 − 6x − 6y + 14 = 0 and also touches Y-axis, is given by the equation- a)x2−6x−10y+14=0
- b)x2−10x−6y+14=0
- c)y2−6x−10y+14=0
- d)y2−10x−6y+14=0
Correct answer is option 'D'. Can you explain this answer?
The locus of the centre of a circle, which touches externally the circle x2 + y2 − 6x − 6y + 14 = 0 and also touches Y-axis, is given by the equation
a)
x2−6x−10y+14=0
b)
x2−10x−6y+14=0
c)
y2−6x−10y+14=0
d)
y2−10x−6y+14=0
|
Ameya Tiwari answered |
Understanding the Circle Equation
The given circle is represented by the equation:
- x² + y² - 6x - 6y + 14 = 0
This can be converted into standard form. Completing the square for both x and y:
- (x - 3)² + (y - 3)² = 4
This represents a circle centered at (3, 3) with a radius of 2.
Locus of the Center of the New Circle
To find the locus of the center of a circle that touches this given circle externally and the Y-axis, consider the following:
- Let the center of the new circle be at (h, k).
- The radius of this new circle is denoted as r.
Conditions for the New Circle
1. Touching the Y-axis: The distance from the center (h, k) to the Y-axis must equal the radius r. Hence, we have:
- r = h
2. Touching the Given Circle Externally: The distance between the centers of the two circles must equal the sum of their radii:
- Distance = √[(h - 3)² + (k - 3)²] = r + 2
Substituting r = h into the distance formula gives:
- √[(h - 3)² + (k - 3)²] = h + 2
Squaring both sides and rearranging leads to the equation of the locus.
Final Equation
After simplification, the resulting equation of the locus of the center (h, k) is:
- y² - 10x - 6y + 14 = 0
This corresponds to option 'D'.
Conclusion
Hence, the locus of the center of the new circle touching both the given circle externally and the Y-axis is described by the equation:
- y² - 10x - 6y + 14 = 0
The given circle is represented by the equation:
- x² + y² - 6x - 6y + 14 = 0
This can be converted into standard form. Completing the square for both x and y:
- (x - 3)² + (y - 3)² = 4
This represents a circle centered at (3, 3) with a radius of 2.
Locus of the Center of the New Circle
To find the locus of the center of a circle that touches this given circle externally and the Y-axis, consider the following:
- Let the center of the new circle be at (h, k).
- The radius of this new circle is denoted as r.
Conditions for the New Circle
1. Touching the Y-axis: The distance from the center (h, k) to the Y-axis must equal the radius r. Hence, we have:
- r = h
2. Touching the Given Circle Externally: The distance between the centers of the two circles must equal the sum of their radii:
- Distance = √[(h - 3)² + (k - 3)²] = r + 2
Substituting r = h into the distance formula gives:
- √[(h - 3)² + (k - 3)²] = h + 2
Squaring both sides and rearranging leads to the equation of the locus.
Final Equation
After simplification, the resulting equation of the locus of the center (h, k) is:
- y² - 10x - 6y + 14 = 0
This corresponds to option 'D'.
Conclusion
Hence, the locus of the center of the new circle touching both the given circle externally and the Y-axis is described by the equation:
- y² - 10x - 6y + 14 = 0
The locus of mid–point of family of chords λx + y - 5 = 0 (parameter) of the parabola x2 = 20y is- a)y2 = 10(x - 5)
- b)x2 = 10(y - 5)
- c)x2 + y2 = 25
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The locus of mid–point of family of chords λx + y - 5 = 0 (parameter) of the parabola x2 = 20y is
a)
y2 = 10(x - 5)
b)
x2 = 10(y - 5)
c)
x2 + y2 = 25
d)
none of these
|
Parth Majumdar answered |
λx + y - 5 = 0 is the focal chord
Since it passes through (0, 5)
Let h, k be the mid point of all such chord h = 5(t1 + t2),
On eliminating t1, t2
x2 = 10 (y - 5).
Since it passes through (0, 5)
Let h, k be the mid point of all such chord h = 5(t1 + t2),
On eliminating t1, t2
x2 = 10 (y - 5).
The line x + y = 6 is a normal to the parabola y2 = 8x at the point- a)(18, - 12)
- b)(4, 2)
- c)(2, 4)
- d)(3, 3)
Correct answer is option 'C'. Can you explain this answer?
The line x + y = 6 is a normal to the parabola y2 = 8x at the point
a)
(18, - 12)
b)
(4, 2)
c)
(2, 4)
d)
(3, 3)
|
Devika Banerjee answered |
Method to Solve :
From the given equation of the parabola, we can find the focus to be4a = 8, this implies, a = 2
The condition for a line y = mx + c, to be a tangent to the parabola, then c = a/m
The given line is x - y +2 = 0
or y = x+2, where m = 1 and c = 2
c = a/m = 2/1 satifies the condition for the line to be a tangent.
Hence the given line is a tangent to the parabola.
The condition for a line y = mx + c, to be a tangent to the parabola, then c = a/m
The given line is x - y +2 = 0
or y = x+2, where m = 1 and c = 2
c = a/m = 2/1 satifies the condition for the line to be a tangent.
Hence the given line is a tangent to the parabola.
t ∈ R represents
The equations x = at2, y = 4at ; t ∈ R represent- a)a parabola
- b)a circle
- c)an ellipse
- d)a hyperbola
Correct answer is option 'A'. Can you explain this answer?
The equations x = at2, y = 4at ; t ∈ R represent
a)
a parabola
b)
a circle
c)
an ellipse
d)
a hyperbola
|
|
Abhay Mehta answered |
The given equations are:
x = at^2
y = 4at
To eliminate the variable t and find an equation relating x and y, we can solve the second equation for t:
t = y / (4a)
Substituting this value of t into the first equation, we get:
x = a(y / (4a))^2
x = (y^2) / (16a)
Therefore, the equation relating x and y is:
x = (y^2) / (16a)
x = at^2
y = 4at
To eliminate the variable t and find an equation relating x and y, we can solve the second equation for t:
t = y / (4a)
Substituting this value of t into the first equation, we get:
x = a(y / (4a))^2
x = (y^2) / (16a)
Therefore, the equation relating x and y is:
x = (y^2) / (16a)
The line y = c is a tangent to the parabola 7/2 if c is equal to- a)a
- b)0
- c)2a
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
The line y = c is a tangent to the parabola 7/2 if c is equal to
a)
a
b)
0
c)
2a
d)
none of these
|
|
Sravya Nair answered |
y = x is tangent to the parabola
y=ax2+c
if a= then c=?
y′ =2ax
y’ = 2(7/2)x =1
x = 1/7
1/7 = 2(1/7)2 + c
c = 1/7 * 2/49
c = 7/2
y=ax2+c
if a= then c=?
y′ =2ax
y’ = 2(7/2)x =1
x = 1/7
1/7 = 2(1/7)2 + c
c = 1/7 * 2/49
c = 7/2
If the tangents at two points (1, 2) and (3, 6) as a parabola intersect at the point (– 1, 1), then the slope of the directrix of the parabola is - a)√2
- b)-2
- c)-1
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
If the tangents at two points (1, 2) and (3, 6) as a parabola intersect at the point (– 1, 1), then the slope of the directrix of the parabola is
a)
√2
b)
-2
c)
-1
d)
none of these
|
Manisha Choudhary answered |
If the tangents at P and Q intersect at T, then axis of parabola is parallel to TR, where R is the mid point of P and Q. So, slope of the axis is 1.
∴ slope of the directrix = – 1.
∴ slope of the directrix = – 1.
If (h, k) is a point on the axis of parabola 2(x -1)2 + 2 (y -1)2 = (x + y + 2)2 from where three distinct normals can be drawn, then- a)h > 2
- b)h < 4
- c)h > 8
- d)h < 8
Correct answer is option 'A'. Can you explain this answer?
If (h, k) is a point on the axis of parabola 2(x -1)2 + 2 (y -1)2 = (x + y + 2)2 from where three distinct normals can be drawn, then
a)
h > 2
b)
h < 4
c)
h > 8
d)
h < 8
|
|
Shounak Shah answered |
which is of the following PM2 = SP2∴ Focus= (1,1) . Directrix is x+y+2 =0
Axis is x-y=0 and z= (-1,-1)
Vertex = (0,0). Parameter a = √2
The distance of the point from vertex and lie on axis from which 3 normals can be drawn must be greater 2a = 2√2. Hence point on axis at a distance 2√2 is (2,2), Hence h>2
A and B are two distinct points, Locus of a point P satisfying |PA| + |PB| = 2k, a constant is- a)nothing can be said
- b)the line segment [AB]
- c)an ellipse
- d)a hyperbola
Correct answer is option 'A'. Can you explain this answer?
A and B are two distinct points, Locus of a point P satisfying |PA| + |PB| = 2k, a constant is
a)
nothing can be said
b)
the line segment [AB]
c)
an ellipse
d)
a hyperbola
|
|
Ram Mohith answered |
I think the locus of P is an ellipse because if you take any point on the ellipse then the sum of distance from focii to that
All chords of the parabola y2 = 4x which subtend right angle at the origin are concurrent at the point:- a)(1, 0)
- b)(2, 0)
- c)(3, 0)
- d)(4, 0)
Correct answer is option 'D'. Can you explain this answer?
All chords of the parabola y2 = 4x which subtend right angle at the origin are concurrent at the point:
a)
(1, 0)
b)
(2, 0)
c)
(3, 0)
d)
(4, 0)
|
Akash Shah answered |
Let y = mx+ c be such chord with extremities A and B .
∴ The combined equation of the pair of lines OA and OB is
∴ Coeff of x2 + Coeff of y2 = 0
∴ c = -4 m
∴ The chord equation is y = m (x- 4) .
∴ The combined equation of the pair of lines OA and OB is
∴ Coeff of x2 + Coeff of y2 = 0
∴ c = -4 m
∴ The chord equation is y = m (x- 4) .
The tangent and normal at the point P(4, 4) to the parabola, y2 = 4x intersect the x–axis at the points Q and R respectively. Then the cirucm centre of the ΔPQR is- a)(2, 0)
- b)(2, 1)
- c)(1, 0)
- d)(1, 2)
Correct answer is option 'C'. Can you explain this answer?
The tangent and normal at the point P(4, 4) to the parabola, y2 = 4x intersect the x–axis at the points Q and R respectively. Then the cirucm centre of the ΔPQR is
a)
(2, 0)
b)
(2, 1)
c)
(1, 0)
d)
(1, 2)
|
Pragati Patel answered |
Eq. of tangent 2y = x + 4
∴ Q ≡ (-4,0)
Eq. of normal is y - 4 = -2 (x - 4)
⇒ y + 2x = 12
Clearly QR is diameter of the required circle.
⇒ (x + 4) (x – 6) + y2 = 0
⇒ x2 + y2 – 2x – 24 = 0
centre (1, 0)
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