Parabola — Page 5 | JEE Mathematics

Q41

If P is a point on the parabola y = x2 + 4 which is closest to the straight line y = 4x

-

1, then the co-ordinates of P are :

A (

-

2, 8)

B (2, 8)

C (1, 5)

D (3, 13)

Correct Answer

Option B

Solution

Given, curve y = x2 + 4 and, line y = 4x $-$ 1 Here, y = x2 + 4 $\therefore$

{{dy} \over {dx}} = 2x

..... (i) and y = 4x $-$ 1

{{dy} \over {dx}} = 4

..... (ii) Let the required point be P(x1, y1). $\therefore$

{\left. {{{dy} \over {dx}}} \right|_P} = 2{x_1}

..... (iii) $\because$ Slopes will be equal. $\therefore$ 2x1 = 4 [from Eqs. (ii) and (iii)]

\Rightarrow {x_1} = {4 \over 2} = 2

Now, the given point P(x1, y1) lies on curve y = x2 + 4, $\therefore$ y1 = x

_1^2

+ 4 $\Rightarrow$ y1 = 22 + 4 = 8 Hence, required coordinate of P = (2, 8)

Q42

The shortest distance between the line x

-

y = 1 and the curve x2 = 2y is :

A 0

B

{1 \over 2{\sqrt 2 }}

C

{1 \over {\sqrt 2 }}

D

{1 \over 2}

Correct Answer

Option B

Solution

Shortest distance must be along common normal Let a point on curve has x coordinate = h Then y coordinate : h2 = 2y $\Rightarrow$ y =

{{{h^2}} \over 2}

So point is = (h,

{{{h^2}} \over 2}

) m1 (slope of line x $-$ y = 1) = 1 $\Rightarrow$ slope of perpendicular line = $-$ 1 Slope of the perpendicular line on the curve x2 = 2y,

{m_2} = {{2x} \over 2} = x \Rightarrow {m_2} = h

$\therefore$ Slope of normal =

- {1 \over h}

- {1 \over h}

= $-$ 1 $\Rightarrow$ h = 1 So point is

\left( {1,{1 \over 2}} \right)

D = \left| {{{1 - {1 \over 2} - 1} \over {\sqrt {1 + 1} }}} \right| = {1 \over {2\sqrt 2 }}

Q43

If the three normals drawn to the parabola, y2 = 2x pass through the point (a, 0) a

\ne

0, then 'a' must be greater than :

A

{1 \over 2}

B 1

C

-

1

D

-

{1 \over 2}

Correct Answer

Option B

Solution

Let the equation of the normal is y = mx $-$ 2am $-$ am3 here 4a = 2 $\Rightarrow$ a =

{1 \over 2}

y = mx $-$ m $-$

{1 \over 2}

m3 It passing through A(a, 0) then 0 = am $-$ m $-$

{1 \over 2}

m3 m = 0, a $-$ 1 $-$

{1 \over 2}

m2 = 0 m2 = 2(a $-$ 1) > 0 $\Rightarrow$ a > 1

Q44

Let C be the locus of the mirror image of a point on the parabola y2 = 4x with respect to the line y = x. Then the equation of tangent to C at P(2, 1) is :

A x

-

y = 1

B 2x + y = 5

C x + 3y = 5

D x + 2y = 4

Correct Answer

Option A

Solution

Image of y2 = 4x w.r.t. y = x is x2 = 4y tangent from (2, 1) xx1 = 2(y + y1) 2x = 2(y + 1) x = y + 1

Q45

Let L be a tangent line to the parabola y2 = 4x

-

20 at (6, 2). If L is also a tangent to the ellipse

{{{x^2}} \over 2} + {{{y^2}} \over b} = 1

, then the value of b is equal to :

A 20

B 14

C 16

D 11

Correct Answer

Option B

Solution

Parabola y2 = 4x $-$ 20 Tangent at P(6, 2) will be

2y = 4\left( {{{x + 6} \over 2}} \right) - 20

2y = 2x + 12 $-$ 20 2y = 2x $-$ 8 y = x $-$ 4 x $-$ y $-$ 4 = 0 ....... (1) This is also tangent to ellipse

{{{x^2}} \over 2} + {{{y^2}} \over b} = 1

Apply c2 = a2m2 + b2 ( $-$ 4)2 = (2)(1) + b b = 14

Q46

Let the tangent to the parabola S : y2 = 2x at the point P(2, 2) meet the x-axis at Q and normal at it meet the parabola S at the point R. Then the area (in sq. units) of the triangle PQR is equal to :

A

{{25} \over 2}

B

{{35} \over 2}

C

{{15} \over 2}

D 25

Correct Answer

Option A

Solution

Tangent at P : y(2) = 2(1/2) (x + 2) $\Rightarrow$ 2y = x + 2 $\therefore$ Q = ( $-$ 2, 0) Normal at P : y $-$ 2 = $-$

{{(2)} \over {2.{1 \over 2}}}(x - 2)

$\Rightarrow$ y $-$ 2 = $-$ 2(x $-$ 2) $\Rightarrow$ y = 6 $-$ 2x $\therefore$ Solving with y2 = 2x $\Rightarrow$ R

\left( {{9 \over 2} - 3} \right)

$\therefore$ Ar (

\Delta

PQR) =

{1 \over 2}\left| \begin{array}{lll}2 & 2 & 1 \\ { - 2} & 1 & 1 \\ {{9 \over 2}} & 3 & { - 1} \end{array} \right|

= {{25} \over 2}

sq. units.

Q47

Let P be a variable point on the parabola

y = 4{x^2} + 1

. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is :

A

{(3x - y)^2} + (x - 3y) + 2 = 0

B

2{(3x - y)^2} + (x - 3y) + 2 = 0

C

{(3x - y)^2} + 2(x - 3y) + 2 = 0

D

2{(x - 3y)^2} + (3x - y) + 2 = 0

Correct Answer

Option B

Solution

Given, parabola

y = 4{x^2} + 1

Let R(a, b) be mid-point of line joining point P and Q where PQ is perpendicular to line y = x.

Let coordinates of P be P(x, y), Q(q, q) and R(a, b) then,

a = {{x + q} \over 2}

and

b = {{y + q} \over 2}

Now, slope of line y = x is m1 = 1 Slope of line PQ be

{{b - q} \over {a - q}} = {m_2}

(say) $\because$ Line y = x and PQ are perpendicular to each other, m1 . m2 = $-$ 1

\Rightarrow {{b - q} \over {a - q}} = - 1 \Rightarrow b - q = q - a

\Rightarrow q = {{b + a} \over 2}

$\therefore$

a = {{x + q} \over 2} = {{x + \left( {{{b + a} \over 2}} \right)} \over 2} = {{2x + b + a} \over 4}

\Rightarrow x = {{4a - b - a} \over 2} = {{3a - b} \over 2}

and

b = {{y + q} \over 2} = {{y + \left( {{{b + a} \over 2}} \right)} \over 2} = {{2y + b + a} \over 4}

\Rightarrow y = {{3b - a} \over 2}

Put (x, y) in equation of parabola as P(x, y) is variable point on parabola

{{3b - a} \over 2} = 4{\left( {{{3a - b} \over 2}} \right)^2} + 1

{{(3b - a)} \over 2} = {(3a - b)^2} + 1

\Rightarrow (3b - a) = 2{(3a - b)^2} + 2

Replace (a, b) as (x, y) $\Rightarrow$ (3y $-$ x) = 2(3x $-$ y)2 + 2 or

2{(3x - y)^2} + (x - 3y) + 2 = 0

Q48

Let a parabola b be such that its vertex and focus lie on the positive x-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from O(0, 0) to the parabola P which meet P at S and R, then the area (in sq. units) of

\Delta

SOR is equal to :

A

16\sqrt 2

B 16

C 32

D

8\sqrt 2

Correct Answer

Option B

Solution

Clearly RS is latus-rectum $\because$ VF = 2 = a $\therefore$ RS = 4a = 8 Now OF = 2a = 4 $\Rightarrow$ Area of triangle ORS = 16

Q49

A tangent and a normal are drawn at the point P(2,

-

4) on the parabola y2 = 8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is equal to :

A

-

16

B

-

18

C

-

12

D

-

20

Correct Answer

Option A

Solution

Given, parabola

{y^2} = 8x

...... (i) Equation of tangent at

P(2, - 4)

is

- 4y = 4(x + 2)

or,

x + y + 2 = 0

..... (ii) and Equation of normal to the parabola is

x - y + C = 0

$\therefore$ Normal passes through

(2, - 4)

$\therefore$

C = - 6

Normal :

x - y = 6

..... (iii) Equation of directrix of parabola

x = - 2

..... (iv) Point of intersection of tangent and normal with directrix are

x = - 2

at

A( - 2,0)

and

B( - 2, - 8)

respectively.

Q(a,b)

and

P(2, - 4)

are given and AQBP is a square. Mid-point of AB = Mid-point of PQ

\Rightarrow ( - 2, - 4) = \left( {{{a + 2} \over 2},{{b - 4} \over 2}} \right) \Rightarrow a = - 6,b = - 4

\Rightarrow 2a + b = - 16

Q50

If two tangents drawn from a point P to the parabola y2 = 16(x

-

3) are at right angles, then the locus of point P is :

A x + 3 = 0

B x + 1 = 0

C x + 2 = 0

D x + 4 = 0

Correct Answer

Option B

Solution

Locus is directrix of parabola x $-$ 3 + 4 = 0 $\Rightarrow$ x + 1 = 0.