Parabola — Page 8 | JEE Mathematics

Q71

Let

A

be a point on the

x

-axis. Common tangents are drawn from

A

to the curves

x^2+y^2=8

and

y^2=16 x

. If one of these tangents touches the two curves at

Q

and

R

, then

(Q R)^2

is equal to :

A 76

B 81

C 64

D 72

Correct Answer

Option D

Solution

Let a tangent on

{y^2} = 16x

be

y = mx + {4 \over m}

For common to

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

{4 \over m} = 2\sqrt 2 (1 + {m^2})

\Rightarrow {2 \over {{m^2}}} = 1 + {m^2} \Rightarrow m = \, \pm 1

Taking one of the tangent

y = x + 4

Point of tangency with

{y^2} = 4x

{x^2} + 8 + 16 = 4x \Rightarrow x = 4

&

y = 8

$\therefore$

Q(4,8)

and for

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

2{x^2} + 8x + 8 = 0

{x^2} + 4x + 4 = 8 \Rightarrow x = - 2,y = 2 \Rightarrow R = ( - 2,2)

{(QR)^2} = {6^2} + {6^2} = 72

Q72

The parabolas :

a x^2+2 b x+c y=0

and

d x^2+2 e x+f y=0

intersect on the line

y=1

. If

a, b, c, d, e, f

are positive real numbers and

a, b, c

are in G.P., then :

A

\dfrac{d}{a}, \dfrac{e}{b}, \dfrac{f}{c}

are in A.P.

B

\dfrac{d}{a}, \dfrac{e}{b}, \dfrac{f}{c}

are in G.P.

C

d, e, f

are in A.P.

D

d, e, f

are in G.P.

Correct Answer

Option A

Solution

Let point of intersection be (

\alpha,1

)

\alpha x^2+2b\alpha+c=0

..... (i) and

d\alpha^2+2e\alpha+f=0

.... (ii)

\Rightarrow a\alpha^2+2\sqrt{ac}\alpha+c=0

( $\because$

b^2=ac

)

{\left( {\sqrt a \alpha + \sqrt c } \right)^2} = 0

\alpha = - \sqrt {{c \over a}}

Put the value of $\alpha$ in (ii),

d{c \over a} - 2e\sqrt {{c \over a}} + f = 0

{d \over a} - {{2e} \over {\sqrt {ac} }} + {f \over c} = 0

{d \over a} + {f \over c} = 2{e \over b}

{d \over a},{e \over b},{f \over c}

are in A.P.

Q73

If

\mathrm{P}(\mathrm{h}, \mathrm{k})

be a point on the parabola

x=4 y^{2}

, which is nearest to the point

\mathrm{Q}(0,33)

, then the distance of

\mathrm{P}

from the directrix of the parabola

\quad y^{2}=4(x+y)

is equal to :

A 8

B 2

C 6

D 4

Correct Answer

Option C

Solution

Equation of normal

y = - tx + 2.{1 \over {16}}t + {1 \over {16}}{t^3}

33 = {t \over 8} + {{{t^3}} \over {16}}

\Rightarrow {t^3} + 2t = 528

t = 8

(a{t^2},2at) = (4,1)

Distance from

x = - 2

Q74

If the tangent at a point P on the parabola

y^2=3x

is parallel to the line

x+2y=1

and the tangents at the points Q and R on the ellipse

\dfrac{x^2}{4}+\dfrac{y^2}{1}=1

are perpendicular to the line

x-y=2

, then the area of the triangle PQR is :

A

\dfrac{9}{\sqrt5}

B

3\sqrt5

C

5\sqrt3

D

\dfrac{3}{2}\sqrt5

Correct Answer

Option B

Solution

P \equiv \left( {{A \over {{m^2}}},{{2A} \over m}} \right)

where

\left( {A = {3 \over 4},m = {{ - 1} \over 2}} \right)

&

Q,R = \left( { \mp \,{{{a^2}{m_1}} \over {{a^2}m_1^2 + {b^2}}},{{ \mp \,.\,{b^2}} \over {\sqrt {{a^2}m_1^2 + {b^2}} }}} \right)

Where

{a^2} = 4,{b^2} = 1

and

{m_1} = 1

$\therefore$

P \equiv (3, - 3)

Q \equiv \left( {{{ - 4} \over {\sqrt 5 }},{{ - 1} \over {\sqrt 5 }}} \right)

&

R\left( {{4 \over {\sqrt 5 }},{1 \over {\sqrt 5 }}} \right)

Area

= {1 \over 2}\left| \begin{array}{lll}3 & { - 3} & 1 \\ {{{ - 4} \over {\sqrt 5 }}} & {{{ - 1} \over {\sqrt 5 }}} & 1 \\ {{4 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} & 1 \end{array} \right| = {1 \over {10}}\left| \begin{array}{lll}3 & { - 3} & 1 \\ { - 4} & { - 1} & {\sqrt 5 } \\ 0 & 0 & {2\sqrt 5 } \end{array} \right|

= {{2\sqrt 5 } \over {10}}( - 15) = 3\sqrt 5

Q75

The equations of two sides of a variable triangle are

x=0

and

y=3

, and its third side is a tangent to the parabola

y^2=6x

. The locus of its circumcentre is :

A

4{y^2} - 18y - 3x - 18 = 0

B

4{y^2} + 18y + 3x + 18 = 0

C

4{y^2} - 18y + 3x + 18 = 0

D

4{y^2} - 18y - 3x + 18 = 0

Correct Answer

Option C

Solution

Third side of triangle $t y=x+\dfrac{3}{2} t^{2}$

\begin{aligned} & \therefore H \equiv(0,3) \quad G \equiv\left(t-\frac{3 t}{2}\right) \end{aligned}

Let O (h, k)

\begin{aligned} & \Rightarrow \frac{2 h}{3}=t-\frac{t^{2}}{2} \& \frac{2 k+3}{3}=2+\frac{t}{2} \\\\ & \Rightarrow 4 h=6 t-3 t^{2} \& 4 k=6+3 t \\\\ & \Rightarrow 4 h=2(4 k-6)-3\left(\frac{(4 k-6)^{2}}{9}\right) \\\\ & \Rightarrow 4 h=6 k-9-\left(4 k^{2}+9-12 k\right) \\\\ & \Rightarrow 4 k^{2}-18 k+3 h+18=0 \\\\ & \Rightarrow 4 y^{2}-18 y+3 x+18=0 \end{aligned}

Q76

The distance of the point

(6,-2\sqrt2)

from the common tangent

\mathrm{y=mx+c,m > 0}

, of the curves

x=2y^2

and

x=1+y^2

is :

A

\dfrac{1}{3}

B 5

C

\dfrac{14}{3}

D 5

\sqrt3

Correct Answer

Option B

Solution

\begin{aligned} & y^2=\frac{x}{2} \Rightarrow \text{ tangent } y=m x+\frac{1}{8 m} \\\\ & y^2=x-1 \Rightarrow \text{ tangent } y=m(x-1)+\frac{1}{4 m} \\\\ & \text{ For common tangent } \frac{1}{8 m}=-m+\frac{1}{4 m} \\\\ & \Rightarrow 1=-8 m^2+2 \\\\ & \because m>0 \Rightarrow m=\frac{1}{2 \sqrt{2}} \\\\ & \Rightarrow \text{ Common tangent is } y=\frac{x}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}} \\\\ & \Rightarrow x-2 \sqrt{2} y+1=0 \end{aligned}

Distance of point $(6,-2 \sqrt{2})$ from common tangent $=5$

Q77

The equations of the sides AB and AC of a triangle ABC are

(\lambda+1)x+\lambda y=4

and

\lambda x+(1-\lambda)y+\lambda=0

respectively. Its vertex A is on the y-axis and its orthocentre is (1, 2). The length of the tangent from the point C to the part of the parabola

y^2=6x

in the first quadrant is :

A 4

B 2

\sqrt2

C 2

D

\sqrt6

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{AB}:(\lambda+1) x+\lambda y=4 \\\\ & \mathrm{AC}: \lambda x+(1-\lambda) y+\lambda=0 \\\\ & \text{ Vertex } A \text{ is on } y \text{-axis } \\\\ & \Rightarrow x=0 \end{aligned}

So $\mathrm{y}=\dfrac{4}{\lambda}, \mathrm{y}=\dfrac{\lambda}{\lambda-1}$

\begin{aligned} & \Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1} \\\\ & \Rightarrow \lambda=2 \\\\ & \mathrm{AB}: 3 x+2 y=4 \\\\ & \mathrm{AC}: 2 \mathrm{x}-\mathrm{y}+2=0 \\\\ & \Rightarrow \mathrm{A}(0,2) \text{ Let } \mathrm{C}(\alpha, 2 \alpha+2) \end{aligned}

Now (Slope of Altitude through C) $\left(-\dfrac{3}{2}\right)=-1$

\left(\frac{2 \alpha}{\alpha-1}\right)\left(-\frac{3}{2}\right)=-1 \Rightarrow \alpha=-\frac{1}{2}

So $\mathrm{C}\left(-\dfrac{1}{2}, 1\right)$ Let Equation of tangent be $y=m x+\dfrac{3}{2 m}$

\begin{aligned} & \mathrm{m}^2+2 \mathrm{~m}-3=0 \\\\ & \Rightarrow \mathrm{m}=1,-3 \end{aligned}

So tangent which touches in first quadrant at $\mathrm{T}$ is

\begin{aligned} & \mathrm{T} \equiv\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right) \\\\ & \equiv\left(\frac{3}{2}, 3\right) \\\\ & \Rightarrow \mathrm{CT}=\sqrt{4+4}=2 \sqrt{2} \end{aligned}

Q78

Let a tangent to the curve

\mathrm{y^2=24x}

meet the curve

xy = 2

at the points A and B. Then the mid points of such line segments AB lie on a parabola with the :

A length of latus rectum 2

B directrix 4x =

-

3

C directrix 4x = 3

D length of latus rectum

\dfrac{3}{2}

Correct Answer

Option C

Solution

Given parabola

{y^2} = 24x = 4.6x = 4ax

$\therefore$ a = 6 If

y = mx + c

is a tangent to the parabola then,

c = {a \over m}

\Rightarrow c = {6 \over m}

Let midpoint of chord AB is

({x_1},{y_1})

We know, locus of midpoint of chord for hyperbola,

T = {S_1}

\Rightarrow {{x{y_1} + {x_1}y} \over 2} - 2 = {x_1}{y_1} - 2

\Rightarrow x{y_1} + {x_1}y = 2{x_1}{y_1}

\Rightarrow y = \left( { - {{{y_1}} \over {{x_1}}}} \right)x + 2{y_1}

This equation is the same tangent to the parabola. Then comparing with

y = mx + c

we get,

c = 2{y_1},\,m = - {{{y_1}} \over {{x_1}}}

$\therefore$

2{y_1} = {6 \over { - {{{y_1}} \over {{x_1}}}}}

\Rightarrow - y_1^2 = 3{x_1}

$\therefore$ Locus of midpoint

({x_1},{y_1})

is,

- {y^2} = 3x

$\therefore$ Length of latus rectum = 3 Equation of directrix is $x=\dfrac{3}{4}$

Q79

Let

\mathrm{PQ}

be a focal chord of the parabola

y^{2}=36 x

of length 100 , making an acute angle with the positive

x

-axis. Let the ordinate of

\mathrm{P}

be positive and

\mathrm{M}

be the point on the line segment PQ such that PM : MQ = 3 : 1. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line

\mathrm{PQ}

?

A

(6,29)

B

(-3,43)

C

(3,33)

D

(-6,45)

Correct Answer

Option B

Solution

The given parabola is of the form

y^2 = 4ax

. Here, 4a = 36, which means a = 9. Length of focal chord at $(t)=a\left(t+\dfrac{1}{t}\right)^2=100$ Where $a=9$

\begin{gathered} t+\frac{1}{t}= \pm \frac{10}{3} \\\\ \therefore \quad t=3, \frac{1}{3},-3, \frac{-1}{3} \end{gathered}

Since ordinate of $P$ is $+\mathrm{ve}$

\therefore t=3 \text{ or } \frac{1}{3}

For each value of t, we can find the coordinates of the points P and Q on the parabola using the parametric form of the parabola $y^2=4ax$ , where a is 9.

The parametric form is given by $x=at^2$ and $y=2at$ .

Let's consider $t=3$ first.

For $t=3$ , the coordinates of P are given by $P(at^2, 2at) = P(81, 54)$ .

Since $t_1t_2=-1$ for a focal chord, the parameter value for Q is $t_2=-1/t_1=-1/3$ .

So the coordinates of Q are given by $Q(at_2^2, 2at_2) = Q(1, -6)$ .

Now, let's find the coordinates of the point M that divides the line segment PQ in the ratio 3:1.

The x-coordinate of M is given by $\dfrac{3Q_x+P_x}{4} = \dfrac{3*1+81}{4} = 21$ .

The y-coordinate of M is given by $\dfrac{3Q_y+P_y}{4} = \dfrac{3*(-6)+54}{4} = 9$ .

So, the coordinates of M are M $(21, 9)$ .

We can repeat these steps for $t=1/3$ to find the other set of points P, Q, and M.

However, since the ordinate of P is positive and PQ makes an acute angle with the positive x-axis, $t=3$ is the appropriate choice in this context.

The slope of the line PQ is $m_{PQ} = \dfrac{-6 - 54}{1 - 81} = \dfrac{-60}{-80} = \dfrac{3}{4}$ .

The slope of the line perpendicular to PQ is $m_{\perp PQ} = -1/m_{PQ} = -\dfrac{4}{3}$ .

Thus, the equation of the line through M and perpendicular to PQ is $(y - 9) = -\dfrac{4}{3}(x - 21)$ , or $4x + 3y = 111$ .

Now, we check which of the given points do NOT lie on this line.

Option A : $(6, 29)$ Substitute these values into the equation : $4\times6 + 3\times29 = 111?$ $24 + 87 = 111?$ $111 = 111$ So, option A does lie on the line.

Option B : $(-3, 43)$ Substitute these values into the equation : $4\times(-3) + 3\times43 = 111?$ $-12 + 129 = 111?$ $117 = 111$ So, option B does NOT lie on the line.

Option C : $(3, 33)$ Substitute these values into the equation : $4\times3 + 3\times33 = 111?$ $12 + 99 = 111?$ $111 = 111$ So, option C does lie on the line.

Option D : $(-6, 45)$ Substitute these values into the equation : $4\times(-6) + 3\times45 = 111?$ $-24 + 135 = 111?$ $111 = 111$ So, option D does lie on the line.

Therefore, $(-3, 43)$ which does not lie on the line passing through M and perpendicular to the line PQ.

Q80

Let

\mathrm{A}(0,1), \mathrm{B}(1,1)

and

\mathrm{C}(1,0)

be the mid-points of the sides of a triangle with incentre at the point

\mathrm{D}

. If the focus of the parabola

y^{2}=4 \mathrm{ax}

passing through

\mathrm{D}

is

(\alpha+\beta \sqrt{2}, 0)

, where

\alpha

and

\beta

are rational numbers, then

\dfrac{\alpha}{\beta^{2}}

is equal to :

A

\dfrac{9}{2}

B 12

C 6

D 8

Correct Answer

Option D

Solution

\begin{aligned} & \text{ So, } \mathrm{D} \equiv\left(\frac{4}{2+2+2 \sqrt{2}}, \frac{4}{2+2+2 \sqrt{2}}\right) \\\\ & \equiv\left(\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}}\right) \\\\ & =\left(\frac{2}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}, \frac{2}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}\right) \\\\ & \equiv(2-\sqrt{2}, 2-\sqrt{2}) \end{aligned}

\begin{aligned} & \because y^2=4 a x \\\\ & (2-\sqrt{2})^2=4 a(2-\sqrt{2}) \\\\ & \Rightarrow 4 a=2-\sqrt{2} \Rightarrow a=\frac{2-\sqrt{2}}{4} \\\\ & \Rightarrow \frac{1}{2}-\frac{\sqrt{2}}{4}=\alpha+\beta \sqrt{2} \\\\ & \Rightarrow \alpha=\frac{1}{2}, \beta=\frac{-1}{4} \\\\ & \text{ So, } \frac{\alpha}{\beta^2}=\frac{\frac{1}{2}}{\frac{1}{16}}=8 \end{aligned}