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Parabola

JEE Mathematics · 105 questions · Page 1 of 11 · Click an option or "Show Solution" to reveal answer

Q1
If y = mx + c is the normal at a point on the parabola y2 = 8x whose focal distance is 8 units, then ∣c∣\left| c \right| is equal to :
A 232\sqrt 3
B 838\sqrt 3
C 10310\sqrt 3
D 16316\sqrt 3
Correct Answer
Option C
Solution

c = −- 29m −- 9m3 a = 2 Given (at2 −- a)2 + 4a2t2 = 64 ⇒\Rightarrow (a(t2 + 1)) = 8 ⇒\Rightarrow t2 + 1 = 4 ⇒\Rightarrow t2 = 3 ⇒\Rightarrow t =

3\sqrt 3

∴\therefore c = 2at(2 + t2) =

23(5)2\sqrt 3 \left( 5 \right)
∣c∣\left| c \right|

= 10

3\sqrt 3
Q2
The shortest distance between the line y = x and the curve y2 = x – 2 is :
A 7427\over 4 \sqrt2
B 787\over8
C 114211\over 4 \sqrt2
D 2
Correct Answer
Option A
Solution

Slope of the line y = x is 1.

∴\therefore The shortest distance between line y = x and parabola y2 = x – 2 is the distance between line y = x and tangent of parabola having slope 1.

Slope of tangent at point P on the parabola y2 = x – 2 is, 2y

dydx{{dy} \over {dx}}

= 1 ⇒\Rightarrow

dydx{{dy} \over {dx}}

=

12y{1 \over {2y}}

= slope ∴\therefore

12y{1 \over {2y}}

= 1 ⇒\Rightarrow y =

12{1 \over 2}

∴\therefore x coordinate of point P is,

14{1 \over 4}

= x – 2 ⇒\Rightarrow x =

94{9 \over 4}

Shortest distance = length of perpendicular from P on line x – y = 0 =

∣94−12∣12+12{{\left| {{9 \over 4} - {1 \over 2}} \right|} \over {\sqrt {{1^2} + {1^2}} }}

=

742{7 \over {4\sqrt 2 }}
Q3
The tangents at the points A(1,3)A(1,3) and B(1,−1)B(1,-1) on the parabola y2−2x−2y=1y^{2}-2 x-2 y=1 meet at the point PP. Then the area (in unit 2{ }^{2} ) of the triangle PABP A B is :
A 4
B 6
C 7
D 8
Correct Answer
Option D
Solution

Given curve :

y2−2x−2y=1{y^2} - 2x - 2y = 1

. Can be written as

(y−1)2=2(x+1){(y - 1)^2} = 2(x + 1)

And, the given information can be plotted as shown in figure Tangent at

A:2y−x−5=0A:2y - x - 5 = 0

{using

T=0T = 0

} Intersection with

y=1y = 1

is

x=−3x = - 3

Hence, point P is

(−3,1)( - 3,1)

Taking advantage of symmetry Area of

ΔPAB=2×12×(1−(−3))×(3−1)\Delta PAB = 2 \times {1 \over 2} \times (1 - ( - 3)) \times (3 - 1)

= 8 sq. units

Q4
Two common tangents to the circle x2+y2=2a2{x^2} + {y^2} = 2{a^2} and parabola y2=8ax{y^2} = 8ax are :
A x=±(y+2a)x = \pm \left( {y + 2a} \right)
B y=±(x+2a)y = \pm \left( {x + 2a} \right)
C x=±(y+a)x = \pm \left( {y + a} \right)
D y=±(x+a)y = \pm \left( {x + a} \right)
Correct Answer
Option B
Solution

Any tangent to the parabola

y2=8ax{y^2} = 8ax

is

y=mx+2am                ...(i)y = mx + {{2a} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

If

(i)(i)

is a tangent to the circle,

x2+y2=2a2{x^2} + {y^2} = 2{a^2}

then,

2a=±2amm2+1\sqrt {2a} = \pm {{2a} \over {m\sqrt {{m^2} + 1} }}
⇒m2(1+m2)=2\Rightarrow {m^2}\left( {1 + {m^2}} \right) = 2
⇒(m2+2)(m2−1)=0\Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0
⇒m=±1.\Rightarrow m = \pm 1.

So from

(i),(i),
y=±(x+2a).y = \pm \left( {x + 2a} \right).
Q5
The normal at the point(bt12,2bt1)\left( {bt_1^2,2b{t_1}} \right) on a parabola meets the parabola again in the point (bt22,2bt2)\left( {bt_2^2,2b{t_2}} \right), then :
A t2=t1+2t1{t_2} = {t_1} + {2 \over {{t_1}}}
B t2=−t1−2t1{t_2} = -{t_1} - {2 \over {{t_1}}}
C t2=−t1+2t1{t_2} = -{t_1} + {2 \over {{t_1}}}
D t2=t1−2t1{t_2} = {t_1} - {2 \over {{t_1}}}
Correct Answer
Option B
Solution

Equation of the normal to a parabola

y2=4bx{y^2} = 4bx

at point

(bt12,2bt1)\left( {bt_1^2,2b{t_1}} \right)

is

y=−t1x+2bt1+bt13y = - {t_1}x + 2b{t_1} + bt_1^3

As given, it also passes through

(bt22,2bt2)\left( {bt_2^2,2b{t_2}} \right)

then

2bt2=t1bt22+2bt1+bt132b{t_2} = {t_1}bt_2^2 + 2b{t_1} + bt_1^3
2t2−2t1=−t1(t22−t12)2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)
=−t1(t2+t1)(t2−t1)= - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)
⇒2=−t1(t2+t1)\Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)
⇒t2+t1=−2t1\Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}
⇒t2=−t1−2t1\Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}
Q6
If a≠0a \ne 0 and the line 2bx+3cy+4d=02bx+3cy+4d=0 passes through the points of intersection of the parabolas y2=4ax{y^2} = 4ax and x2=4ay{x^2} = 4ay, then :
A d2+(3b−2c)2=0{d^2} + {\left( {3b - 2c} \right)^2} = 0
B d2+(3b+2c)2=0{d^2} + {\left( {3b + 2c} \right)^2} = 0
C d2+(2b−3c)2=0{d^2} + {\left( {2b - 3c} \right)^2} = 0
D d2+(2b+3c)2=0{d^2} + {\left( {2b + 3c} \right)^2} = 0
Correct Answer
Option D
Solution

Solving equations of parabolas

y2=4ax{y^2} = 4ax

and

x2=4ay{x^2} = 4ay

we get

(0,0)(0,0)

and

(4a,4a)(4a, 4a)

Substituting in the given equation of line

2bx+3cy+4d=0,2bx+3cy+4d=0,

we get

d=0d=0

and

2b+3c=02b+3c=0
⇒d2+(2b+3c)2=0\Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0
Q7
The locus of a point which divides the line segment joining the point (0, –1) and a point on the parabola, x2 = 4y, internally in the ratio 1 : 2, is :
A 9x2 – 3y = 2
B 4x2 – 3y = 2
C x2 – 3y = 2
D 9x2 – 12y = 8
Correct Answer
Option D
Solution

Take point P(2t, t2 ) on parabola x2 = 4y h =

2t+03{{2t + 0} \over 3}

and k =

t2−23{{{t^2} - 2} \over 3}

⇒\Rightarrow t =

3h2{{3h} \over 2}

and 3k + 2 = t2 ∴\therefore 3k + 2 =

9h24{{9{h^2}} \over 4}

⇒\Rightarrow 9x2 – 12y = 8

Q8
Let PP be the point (1,0)(1, 0) and QQ a point on the parabola y2=8x{y^2} = 8x. The locus of mid point of PQPQ is :
A y2−4x+2=0{y^2} - 4x + 2 = 0
B y2+4x+2=0{y^2} + 4x + 2 = 0
C x2+4y+2=0{x^2} + 4y + 2 = 0
D x2−4y+2=0{x^2} - 4y + 2 = 0
Correct Answer
Option A
Solution
P=(1,0)  Q=(h,k)P = \left( {1,0} \right)\,\,Q = \left( {h,k} \right)

Such that

k2=8h{k^2} = 8h

Let

(α,β)\left( {\alpha ,\beta } \right)

be the midpoint of

PQPQ
α=h+12,   β=k+02\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}

∴\therefore

2α−1=h      2β=k.2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.
(2β)2=8(2α−1)⇒β2=4α−2{\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2
⇒y2−4x+2=0.\Rightarrow {y^2} - 4x + 2 = 0.
Q9
The locus of the vertices of the family of parabolas y=a3x23+a2x2−2ay = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a is :
A xy=10564xy = {{105} \over {64}}
B xy=34xy = {{3} \over {4}}
C xy=3516xy = {{35} \over {16}}
D xy=64105xy = {{64} \over {105}}
Correct Answer
Option A
Solution

Given parabola is

y=a3x23+a2x2−2ay = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a
⇒y=a33(x3+32a+x+916a2)−3a16−2a\Rightarrow y = {{{a^3}} \over 3}\left( {{x^3} + {3 \over {2a}} + x + {9 \over {16{a^2}}}} \right) - {{3a} \over {16}} - 2a
⇒y+35a16=a33(x+34a)2\Rightarrow y + {{35a} \over {16}} = {{{a^3}} \over 3}{\left( {x + {3 \over {4a}}} \right)^2}

∴\therefore Vertex of parabola is

(−34a,−35a16)\left( {{{ - 3} \over {4a}},{{ - 35a} \over {16}}} \right)

To find locus of this vertex,

x=−34a  x = {{ - 3} \over {4a}}\,\,

and

  y=−35a16\,\,y = {{ - 35a} \over {16}}
⇒a=−34x  \Rightarrow a = {{ - 3} \over {4x}}\,\,

and

a=−16y35a = - {{16y} \over {35}}
⇒−34x=−16y35⇒64xy=105\Rightarrow {{ - 3} \over {4x}} = {{ - 16y} \over {35}} \Rightarrow 64xy = 105
⇒xy=10564\Rightarrow xy = {{105} \over {64}}

which is the required locus.

Q10
The equation of a tangent to the parabola y2=8x{y^2} = 8x is y=x+2y=x+2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is :
A (2,4)(2,4)
B (−2,0)(-2,0)
C (−1,1)(-1,1)
D (0,2)(0,2)
Correct Answer
Option B
Solution

Parabola

y2=8x{y^2} = 8x

We know that the locus of point of intersection of two perpendicular tangents to a parabola is its directrix.

Point must be on the directrix of parabola as equation of directrix

x+2=0⇒x=−2x + 2 = 0 \Rightarrow x = - 2

Hence the point is

(−2,0)\left( { - 2,0} \right)
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