QBManager

Parabola

120986 Equation of the directrix of the conic $x^{2} + 4 y +$ $4 = 0$ is

1

y = 1

2

y = - 1

3

y = 0

4

x = 0

5

x = 1

Explanation:

C We have the equation of conic, $x^{2} + 4 y + 4 = 0$
$x^{2} = - 4 y - 4 = - 4 (y + 1)$
$x^{2} = - 4 Y$
Where $X = x$ and $Y = y + 1$
Equation of directrix is $Y = 1$
$y + 1 = 1$
$∴ y = 0$

Kerala CEE-2008

Parabola

120987 If the vertex of the parabola $y = x^{2} - 16 x + k$ lies on $x$ -axis, then the value of $k$ is :

1 16

2 8

3 64

4 -64

5 -8

Explanation:

C Given, Vertex of the parabola
$y = x^{2} - 16 x + k$
$= x^{2} - 2 x (8) + k$
$= x^{2} - 2 x (8) + 64 - 64 + k$
$= (x - 8)^{2} - 64 + k$
Since, vertex lies on $x$ -axis
$∴ y = 0$ So, it can be possible only when, $x = 8, k = 64$

Kerala CEE-2006

Parabola

120988 An equilateral triangle $SAB$ is inscribed in the parabola $y^{2} = 4 a x$ having its focus at $S$ . It chord $A B$ lies towards the left of $S$ , then side length of this triangle is

1

2 a (2 - \sqrt{3})

2

4 a (2 - \sqrt{3})

3

a (2 - \sqrt{3})

4

8 a (2 - \sqrt{3})

Explanation:

B Let A $({at}_{1}^{2}, 2 {at}_{2}), B ({at}_{2}^{2}, - 2 {at}_{2})$
We have,
$m_{AS} = \tan (\frac{5 π}{6})$
$\frac{2 {at}_{1}}{{at}_{1}^{2} - a} = - \frac{1}{\sqrt{3}}$
$t_{1}^{2} + 2 \sqrt{3} t_{1} - 1 = 0$
$t_{1} = \frac{- 2 \sqrt{3} \pm \sqrt{12 + 4}}{2}$
$t_{1} = \frac{+ 2 (- \sqrt{3} \pm 2)}{2}$
$t_{1} = (- \sqrt{3} \pm 2)$
Clearly, $t_{1} = - \sqrt{3} - 2$ is rejected.
thus, $t_{1} = (2 - \sqrt{3})$
Hence, $AB = 4 {at}_{1}$
$= 4 a (2 - \sqrt{3})$

Manipal UGET-2013

Parabola

120931 The two ends of a latus rectum of a parabola are $(5, 8)$ and $(- 7, 8)$ Then its focus is

1

(- 1, 8)

2

(- 2, 8)

3

(- 2, 16)

4

(- 1, - 8)

Explanation:

A Ends of the lotus rectum are $(5, 8)$ and $(- 7, 8)$ We know focus is the midpoint of two end points of latus rectum.
$So, focus = (\frac{5 - 7}{2}, \frac{8 + 8}{2})$
$= (- 1, 8)$

JEEE-2016

Parabola

120986 Equation of the directrix of the conic $x^{2} + 4 y +$ $4 = 0$ is

1

y = 1

2

y = - 1

3

y = 0

4

x = 0

5

x = 1

Explanation:

C We have the equation of conic, $x^{2} + 4 y + 4 = 0$
$x^{2} = - 4 y - 4 = - 4 (y + 1)$
$x^{2} = - 4 Y$
Where $X = x$ and $Y = y + 1$
Equation of directrix is $Y = 1$
$y + 1 = 1$
$∴ y = 0$

Kerala CEE-2008

Parabola

120987 If the vertex of the parabola $y = x^{2} - 16 x + k$ lies on $x$ -axis, then the value of $k$ is :

1 16

2 8

3 64

4 -64

5 -8

Explanation:

C Given, Vertex of the parabola
$y = x^{2} - 16 x + k$
$= x^{2} - 2 x (8) + k$
$= x^{2} - 2 x (8) + 64 - 64 + k$
$= (x - 8)^{2} - 64 + k$
Since, vertex lies on $x$ -axis
$∴ y = 0$ So, it can be possible only when, $x = 8, k = 64$

Kerala CEE-2006

Parabola

120988 An equilateral triangle $SAB$ is inscribed in the parabola $y^{2} = 4 a x$ having its focus at $S$ . It chord $A B$ lies towards the left of $S$ , then side length of this triangle is

1

2 a (2 - \sqrt{3})

2

4 a (2 - \sqrt{3})

3

a (2 - \sqrt{3})

4

8 a (2 - \sqrt{3})

Explanation:

B Let A $({at}_{1}^{2}, 2 {at}_{2}), B ({at}_{2}^{2}, - 2 {at}_{2})$
We have,
$m_{AS} = \tan (\frac{5 π}{6})$
$\frac{2 {at}_{1}}{{at}_{1}^{2} - a} = - \frac{1}{\sqrt{3}}$
$t_{1}^{2} + 2 \sqrt{3} t_{1} - 1 = 0$
$t_{1} = \frac{- 2 \sqrt{3} \pm \sqrt{12 + 4}}{2}$
$t_{1} = \frac{+ 2 (- \sqrt{3} \pm 2)}{2}$
$t_{1} = (- \sqrt{3} \pm 2)$
Clearly, $t_{1} = - \sqrt{3} - 2$ is rejected.
thus, $t_{1} = (2 - \sqrt{3})$
Hence, $AB = 4 {at}_{1}$
$= 4 a (2 - \sqrt{3})$

Manipal UGET-2013

Parabola

120931 The two ends of a latus rectum of a parabola are $(5, 8)$ and $(- 7, 8)$ Then its focus is

1

(- 1, 8)

2

(- 2, 8)

3

(- 2, 16)

4

(- 1, - 8)

Explanation:

A Ends of the lotus rectum are $(5, 8)$ and $(- 7, 8)$ We know focus is the midpoint of two end points of latus rectum.
$So, focus = (\frac{5 - 7}{2}, \frac{8 + 8}{2})$
$= (- 1, 8)$

JEEE-2016

Parabola

120986 Equation of the directrix of the conic $x^{2} + 4 y +$ $4 = 0$ is

1

y = 1

2

y = - 1

3

y = 0

4

x = 0

5

x = 1

Explanation:

C We have the equation of conic, $x^{2} + 4 y + 4 = 0$
$x^{2} = - 4 y - 4 = - 4 (y + 1)$
$x^{2} = - 4 Y$
Where $X = x$ and $Y = y + 1$
Equation of directrix is $Y = 1$
$y + 1 = 1$
$∴ y = 0$

Kerala CEE-2008

Parabola

120987 If the vertex of the parabola $y = x^{2} - 16 x + k$ lies on $x$ -axis, then the value of $k$ is :

1 16

2 8

3 64

4 -64

5 -8

Explanation:

C Given, Vertex of the parabola
$y = x^{2} - 16 x + k$
$= x^{2} - 2 x (8) + k$
$= x^{2} - 2 x (8) + 64 - 64 + k$
$= (x - 8)^{2} - 64 + k$
Since, vertex lies on $x$ -axis
$∴ y = 0$ So, it can be possible only when, $x = 8, k = 64$

Kerala CEE-2006

Parabola

120988 An equilateral triangle $SAB$ is inscribed in the parabola $y^{2} = 4 a x$ having its focus at $S$ . It chord $A B$ lies towards the left of $S$ , then side length of this triangle is

1

2 a (2 - \sqrt{3})

2

4 a (2 - \sqrt{3})

3

a (2 - \sqrt{3})

4

8 a (2 - \sqrt{3})

Explanation:

B Let A $({at}_{1}^{2}, 2 {at}_{2}), B ({at}_{2}^{2}, - 2 {at}_{2})$
We have,
$m_{AS} = \tan (\frac{5 π}{6})$
$\frac{2 {at}_{1}}{{at}_{1}^{2} - a} = - \frac{1}{\sqrt{3}}$
$t_{1}^{2} + 2 \sqrt{3} t_{1} - 1 = 0$
$t_{1} = \frac{- 2 \sqrt{3} \pm \sqrt{12 + 4}}{2}$
$t_{1} = \frac{+ 2 (- \sqrt{3} \pm 2)}{2}$
$t_{1} = (- \sqrt{3} \pm 2)$
Clearly, $t_{1} = - \sqrt{3} - 2$ is rejected.
thus, $t_{1} = (2 - \sqrt{3})$
Hence, $AB = 4 {at}_{1}$
$= 4 a (2 - \sqrt{3})$

Manipal UGET-2013

Parabola

120931 The two ends of a latus rectum of a parabola are $(5, 8)$ and $(- 7, 8)$ Then its focus is

1

(- 1, 8)

2

(- 2, 8)

3

(- 2, 16)

4

(- 1, - 8)

Explanation:

A Ends of the lotus rectum are $(5, 8)$ and $(- 7, 8)$ We know focus is the midpoint of two end points of latus rectum.
$So, focus = (\frac{5 - 7}{2}, \frac{8 + 8}{2})$
$= (- 1, 8)$

JEEE-2016

Parabola

120986 Equation of the directrix of the conic $x^{2} + 4 y +$ $4 = 0$ is

1

y = 1

2

y = - 1

3

y = 0

4

x = 0

5

x = 1

Explanation:

C We have the equation of conic, $x^{2} + 4 y + 4 = 0$
$x^{2} = - 4 y - 4 = - 4 (y + 1)$
$x^{2} = - 4 Y$
Where $X = x$ and $Y = y + 1$
Equation of directrix is $Y = 1$
$y + 1 = 1$
$∴ y = 0$

Kerala CEE-2008

Parabola

120987 If the vertex of the parabola $y = x^{2} - 16 x + k$ lies on $x$ -axis, then the value of $k$ is :

1 16

2 8

3 64

4 -64

5 -8

Explanation:

C Given, Vertex of the parabola
$y = x^{2} - 16 x + k$
$= x^{2} - 2 x (8) + k$
$= x^{2} - 2 x (8) + 64 - 64 + k$
$= (x - 8)^{2} - 64 + k$
Since, vertex lies on $x$ -axis
$∴ y = 0$ So, it can be possible only when, $x = 8, k = 64$

Kerala CEE-2006

Parabola

120988 An equilateral triangle $SAB$ is inscribed in the parabola $y^{2} = 4 a x$ having its focus at $S$ . It chord $A B$ lies towards the left of $S$ , then side length of this triangle is

1

2 a (2 - \sqrt{3})

2

4 a (2 - \sqrt{3})

3

a (2 - \sqrt{3})

4

8 a (2 - \sqrt{3})

Explanation:

B Let A $({at}_{1}^{2}, 2 {at}_{2}), B ({at}_{2}^{2}, - 2 {at}_{2})$
We have,
$m_{AS} = \tan (\frac{5 π}{6})$
$\frac{2 {at}_{1}}{{at}_{1}^{2} - a} = - \frac{1}{\sqrt{3}}$
$t_{1}^{2} + 2 \sqrt{3} t_{1} - 1 = 0$
$t_{1} = \frac{- 2 \sqrt{3} \pm \sqrt{12 + 4}}{2}$
$t_{1} = \frac{+ 2 (- \sqrt{3} \pm 2)}{2}$
$t_{1} = (- \sqrt{3} \pm 2)$
Clearly, $t_{1} = - \sqrt{3} - 2$ is rejected.
thus, $t_{1} = (2 - \sqrt{3})$
Hence, $AB = 4 {at}_{1}$
$= 4 a (2 - \sqrt{3})$

Manipal UGET-2013

Parabola

120931 The two ends of a latus rectum of a parabola are $(5, 8)$ and $(- 7, 8)$ Then its focus is

1

(- 1, 8)

2

(- 2, 8)

3

(- 2, 16)

4

(- 1, - 8)

Explanation:

A Ends of the lotus rectum are $(5, 8)$ and $(- 7, 8)$ We know focus is the midpoint of two end points of latus rectum.
$So, focus = (\frac{5 - 7}{2}, \frac{8 + 8}{2})$
$= (- 1, 8)$

JEEE-2016