Parabola — Page 9 | JEE Mathematics

Q81

If the line

3 x-2 y+12=0

intersects the parabola

4 y=3 x^2

at the points

A

and

B

, then at the vertex of the parabola, the line segment AB subtends an angle equal to

A

\dfrac{\pi}{2}-\tan ^{-1}\left(\dfrac{3}{2}\right)

B

\tan ^{-1}\left(\dfrac{9}{7}\right)

C

\tan ^{-1}\left(\dfrac{11}{9}\right)

D

\tan ^{-1}\left(\dfrac{4}{5}\right)

Correct Answer

Option B

Solution

\begin{aligned} & 3 x-2 y+12=0 \\ & 4 y=3 x^2 \\ & \therefore 2(3 x+12)=3 x^2 \\ & \Rightarrow x^2-2 x-8=0 \\ & \Rightarrow x=-2,4 \\ & \mathrm{~m}_{\mathrm{OA}}=-3 / 2, \mathrm{~m}_{\mathrm{OB}}=3 \\ & \tan \theta=\left(\frac{\frac{-3}{2}-3}{1-\frac{9}{2}}\right)=\frac{9}{7} \\ & \theta=\tan ^{-1}\left(\frac{9}{7}\right) \text{ (angle will be acute) } \end{aligned}

Q82

Let

R

be the focus of the parabola

y^{2}=20 x

and the line

y=m x+c

intersect the parabola at two points

P

and

Q

. Let the point

G(10,10)

be the centroid of the triangle

P Q R

. If

c-m=6

, then

(P Q)^{2}

is :

A 317

B 325

C 346

D 296

Correct Answer

Option B

Solution

y^2=20 x, y=m x+\mathrm{c}

Put value of $x$

\begin{aligned} & y^2=20\left(\frac{y-c}{m}\right) \\\\ & \Rightarrow y^2-\frac{20}{m} y+\frac{20}{m} c=0 .......(i) \end{aligned}

Since, centroid $=(10,10)$

\begin{aligned} & \text{ So, } \frac{y_1+y_2+0}{3}=10 \\\\ & \Rightarrow y_1+y_2=30 \end{aligned}

From (1),

\text{ Sum of roots }=\frac{20}{m}=30 \Rightarrow m=\frac{2}{3}

Also, $c-m=6 \Rightarrow c=6+\dfrac{2}{3}=\dfrac{20}{3}$ Now, the equation is :

\begin{aligned} & y^2-\frac{20}{2} \times 3 y+\frac{20}{2} \times 3 \times \frac{20}{3}=0 \\\\ & \Rightarrow y^2-30 y+200=0 \\\\ & \Rightarrow y^2-20 y-10 y+200=0 \\\\ & \Rightarrow(y-20)(y-10)=0 \\\\ & \Rightarrow y=10,20 \Rightarrow x=5, x=20 \\\\ & \therefore P \equiv(5,10), Q \equiv(20,20) \\\\ & \text{ So, }(P Q)^2=(20-5)^2+(20-10)^2 \\\\ & =225+100=325 \end{aligned}

Q83

Let

P Q

be a chord of the parabola

y^2=12 x

and the midpoint of

P Q

be at

(4,1)

. Then, which of the following point lies on the line passing through the points

\mathrm{P}

and

\mathrm{Q}

?

A

(3,-3)

B

\left(\dfrac{1}{2},-20\right)

C

(2,-9)

D

\left(\dfrac{3}{2},-16\right)

Correct Answer

Option B

Solution

y^2=12 x

Chord

P Q

having mid-point

(x_1, y_1)=(4,1)

equation of chord

P Q

\begin{aligned} & T=S_1 \\ & y y_1-12 \frac{\left(x+x_1\right)}{2}=y_1^2-12 x_1 \\ & y-6(x+4)=1-12 \times 4 \\ & y-6 x-24=-47 \\ & y-6 x+23=0 \end{aligned}

From option (4)

x=\frac{1}{2}

&

y=-20

-20-6 \times \frac{1}{2}+23=0

Q84

Let

C

be the circle of minimum area touching the parabola

y=6-x^2

and the lines

y=\sqrt{3}|x|

. Then, which one of the following points lies on the circle

C

?

A

(1,2)

B

(2,2)

C

(1,1)

D

(2,4)

Correct Answer

Option D

Solution

Let centre be (0, k) Now radius is

r=6-k

Also,

6-k=\left|\frac{k}{2}\right|

\begin{aligned} & \Rightarrow 6-k=\frac{k}{2} \\ & \Rightarrow 12-2 k=k \\ & \Rightarrow k=4 \end{aligned}

Radius,

r=6-4=2

So circle will be

\begin{aligned} & (x)^2+(y-k)^2=4 \\ & x^2+(y-4)^2=4 \end{aligned}

(2,4)

satisfies this equation.

Q85

Two parabolas have the same focus (4, 3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersect at the points A and B, then (AB)2 is equal to :

A 384

B 392

C 96

D 192

Correct Answer

Option D

Solution

Let intersection points of these two parabolas are

\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right) \& \mathrm{~B}\left(\mathrm{x}_2, \mathrm{y}_2\right)

$\because$ equation of parabola I and II are given below

\begin{aligned} & \therefore(\mathrm{x}-4)^2+(\mathrm{y}-3)^2=\mathrm{x}^2 \quad\text{..... (1)}\\ & \&(\mathrm{x}-4)^2+(\mathrm{y}-3)^2=\mathrm{y}^2 \quad\text{..... (2)} \end{aligned}

Here $A\left(x_1, y_1\right) \& B\left(x_2, y_2\right)$ will satisfy the equation Also from equations (1) & (2), we get $x=y$ .......(

3) Put $x=y$ in equation (1) We get $x^2-14 x+25=0$

\begin{aligned} & x_1+x_2=14 \\ & x_1 x_2=25 \\ & \therefore A B^2=\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2 \\ & =2\left(x_1-x_2\right)^2 \\ & =2\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right] \\ & =192 \end{aligned}

Q86

Let the parabola

y=x^2+\mathrm{p} x-3

, meet the coordinate axes at the points

\mathrm{P}, \mathrm{Q}

and R . If the circle C with centre at

(-1,-1)

passes through the points

P, Q

and

R

, then the area of

\triangle P Q R

is :

A 4

B 6

C 5

D 7

Correct Answer

Option B

Solution

The given parabola is $y = x^2 + px - 3$ .

Intersection with the y-axis: At $x = 0$ , we find $y = -3$ .

Thus, the parabola intersects the y-axis at the point $(0, -3)$ .

Circle Equation: We are given the circle has its center at $(-1, -1)$ and it passes through the points where the parabola intersects the axes.

The radius can be found using the distance from the center to any given point the circle passes through.

Using $(0, -3)$ : $\text{Radius} = \sqrt{(0 + 1)^2 + (-3 + 1)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{5}$ Therefore, the equation of the circle is: $(x + 1)^2 + (y + 1)^2 = 5$ This simplifies to: $x^2 + 2x + 1 + y^2 + 2y + 1 = 5$ $x^2 + y^2 + 2x + 2y + 2 = 5$ $x^2 + y^2 + 2x + 2y - 3 = 0$ Intersection with the x-axis: When $y = 0$ , solving the quadratic $x^2 + 2x - 3 = 0$ gives: $(x + 3)(x - 1) = 0$ So, $x = -3$ or $x = 1$ .

Thus, the intersection points on the x-axis are $(-3, 0)$ and $(1, 0)$ .

Vertices of Triangle $\triangle PQR$ : The vertices of the triangle formed are $P = (0, -3)$ , $Q = (-3, 0)$ , and $R = (1, 0)$ .

Area of $\triangle PQR$ : Use the determinant formula to find the area of the triangle: $\text{Area} = \dfrac{1}{2} \left| \begin{array}{ccc} 0 & -3 & 1 \\ -3 & 0 & 1 \\ 1 & 0 & 1 \end{array} \right|$ Calculate the determinant: $= \dfrac{1}{2} \left( (0)(0)(1) + (-3)(1)(1) + (1)(-3)(1) - (1)(0)(1) - (-3)(1)(0) - (0)(-3)(1) \right)$ Simplify: $= \dfrac{1}{2} \left( 0 - 3 + 3 + 0 \right) = \dfrac{1}{2} (12) = 6$ Thus, the area of $\triangle PQR$ is 6.

Q87

Let

\mathrm{P}(4,4 \sqrt{3})

be a point on the parabola

y^2=4 \mathrm{a} x

and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to :

A

\dfrac{34 \sqrt{3}}{3}

B

\dfrac{343 \sqrt{3}}{8}

C

17 \sqrt{3}

D

\dfrac{263 \sqrt{3}}{8}

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & (4,4 \sqrt{3}) \text{ lies on } \mathrm{y}^2=4 \mathrm{ax} \\ & \Rightarrow 48=4 \mathrm{a} .4 \\ & \quad 4 \mathrm{a}=12 \\ & \Rightarrow \mathrm{y}^2=12 \mathrm{x} \text{ is equation of parabola } \end{aligned}\\ &\text{ Now, parameter of } P \text{ is } t_1=\frac{2}{\sqrt{3}} \Rightarrow \text{ Parameters of } Q \end{aligned}

\begin{aligned} &\text{ is } \mathrm{t}_2=-\frac{\sqrt{3}}{2} \Rightarrow \mathrm{Q}\left(\frac{9}{4},-3 \sqrt{3}\right)\\ &\text{ Area of trapezium PQNM }\\ &\begin{aligned} & =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PM}+\mathrm{QN}) \\ & =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PS}+\mathrm{QS}) \\ & =\frac{1}{2} \mathrm{MN} \cdot \mathrm{PQ} \\ & =\frac{1}{2} 7 \sqrt{3} \cdot \frac{49}{4}=(343) \frac{\sqrt{3}}{8}=3 \end{aligned} \end{aligned}

Q88

The axis of a parabola is the line

y=x

and its vertex and focus are in the first quadrant at distances

\sqrt{2}

and

2 \sqrt{2}

units from the origin, respectively. If the point

(1, k)

lies on the parabola, then a possible value of k is :

A 8

B 3

C 9

D 4

Correct Answer

Option C

Solution

Equation of directrix

\Rightarrow y=-x

By definition of parabola,

\begin{aligned} & P M=P F \\ & \left|\frac{1+K}{\sqrt{2}}\right|=\sqrt{(1-2)^2+(K-2)^2} \\ & \frac{(1+K)^2}{2}=1+K^2+4-4 K \\ & 1+K^2+2 K=10+2 K^2-8 K \\ & K^2-10 K+9=0 \\ & (K-9)(K-1)=0 \\ & \therefore \quad K=1 \text{ or } K=9 \end{aligned}

Q89

Let the shortest distance from

(a, 0), a>0

, to the parabola

y^2=4 x

be 4 . Then the equation of the circle passing through the point

(a, 0)

and the focus of the parabola, and having its centre on the axis of the parabola is :

A

x^2+y^2-8 x+7=0

B

x^2+y^2-6 x+5=0

C

x^2+y^2-4 x+3=0

D

x^2+y^2-10 x+9=0

Correct Answer

Option B

Solution

\begin{aligned} &\text{ Normal at } \mathrm{P}\\ &\begin{gathered} y+t x=2 t+t^3 \\ \uparrow \\ (a, 0) \end{gathered} \end{aligned}

\begin{aligned} & \mathrm{at}=2 \mathrm{t}+\mathrm{t}^3 \\ & \mathrm{a}=2+\mathrm{t}^2 \\ & \mathrm{R}\left(2+\mathrm{t}^2, 0\right) \\ & \mathrm{PR}=4 \Rightarrow 4+4 \mathrm{t}^2=16 \end{aligned}

\begin{aligned} & \quad 4 \mathrm{t}^2=12 \Rightarrow \mathrm{t}^2=3 \\ & \mathrm{a}=5, \mathrm{R}(5,0) \end{aligned}

Focus $(1,0)$ $(1,0) \&(5,0)$ will be the end points of diameter $\Rightarrow E q^{\mathrm{n}}$ of circle is

\begin{aligned} & (x-1)(x-5)+y^2=0 \\ & x^2+y^2-6 x+5=0 \end{aligned}

Q90

If the equation of the parabola with vertex

\mathrm{V}\left(\dfrac{3}{2}, 3\right)

and the directrix

x+2 y=0

is

\alpha x^2+\beta y^2-\gamma x y-30 x-60 y+225=0

, then

\alpha+\beta+\gamma

is equal to :

A 6

B 8

C 7

D 9

Correct Answer

Option D

Solution

Equation of axis

\begin{aligned} & y-3=2\left(x-\frac{3}{2}\right) \\ & y-2 x=0 \end{aligned}

foot of directrix

\begin{aligned} & \quad y-2 x=0 \\ & \& \quad \Rightarrow(0,0)\\ & 2 y+x=0 \end{aligned}

\begin{aligned} & \text{ Focus }=(3,6) \\ & \operatorname{PS}^2=P^2 \\ & (x-3)^2+(y-6)^2=\left(\frac{x+2 y}{\sqrt{5}}\right)^2 \\ & 4 x^2+y^2-4 x y-30 x-60 y+225=0 \\ & \Rightarrow \alpha=4, \beta=1, \gamma=4 \Rightarrow \alpha+\beta+\gamma=9 \end{aligned}