QBManager

Parabola

120953 The distance between the vertex and the focus to the parabola $x^{2} - 2 x - 3 y - 2 = 0$ is

1

\frac{4}{5}

2

\frac{3}{4}

3

\frac{1}{2}

4

\frac{5}{6}

Explanation:

B Given,
$x^{2} - 2 x - 3 y - 2 = 0$
$x^{2} - 2 x + (1)^{2} - (1)^{2} - 3 y - 2 = 0$
$(x - 1)^{2} = 3 y + 3$
$(x - 1)^{2} = 3 (y + 1)$
We know that, distance between vertex and focus of parabola is a.
From equation (i),
$∴ 4 a = 3$
$a = \frac{3}{4}$

AP EAMCET-2016

Parabola

120954 The locus of the mid-points of all chords of the parabola $y^{2} = 4$ ax through its vertex is another parabola with directrix

1

x = - a

2

x = a

3

x = 0

4

x = - \frac{a}{2}

Explanation:

D Given,
$y^{2} = 4 a x$
Any point on parabola (at $^{2}, 2 a t)$
Chord is passed through vertex $(0, 0)$
Midpoint may be $(\frac{{at}^{2}}{2}$ , at $) = (h, k)$
Where, $h = \frac{{at}^{2}}{2}, k = at$
$2 h = a {(\frac{k}{a})}^{2}$
$2 ah = k^{2}$
Replacing $h$ by $x$ and $k$ by $y$ , we get-
$2 a x = y^{2}$
$y^{2} = 4 (\frac{a}{2}) x$ Thus, directrix is, $x = \frac{- a}{2}$

WB JEE-2016

Parabola

120955 The length of the latus rectum of the parabola $20 (x^{2} + y^{2} - 6 x - 2 y + 10) = (4 x - 2 y - 5)^{2}$ , is

1

\frac{\sqrt{5}}{2}

2

2 \sqrt{5}

3

\sqrt{5}

4

4 \sqrt{5}

Explanation:

C Given,
$20 (x^{2} + y^{2} - 6 x - 2 y + 10) = (4 x - 2 y - 5)^{2}$
$\Rightarrow 20 ((x - 3)^{2} + (y - 1)^{2}) = (4 x - 2 y - 5)^{2}$
$Focus = (3, 1)$
$Directrix : 4 x - 2 y - 5 = 0$
$∴ Latus rectum, l = 2 \times | \frac{3 \times 4 - 2 \times 1 - 5}{\sqrt{(4)^{2} + (- 2)^{2}}} |$
$l = 2 \times | \frac{5}{\sqrt{20}} |$
$l = 2 \times \frac{5}{2 \sqrt{5}}$
$l = \sqrt{5}$

AP EAMCET-22.04.2019

Parabola

120956 The equation of the latus rectum of a parabola is $x + y = 8$ and the equation of the tangent at the vertex is $x + y = 12$ . Then, the length of the latus rectum is

1

4 \sqrt{2}

units

2

2 \sqrt{2}

units

3 8 units

4

8 \sqrt{2}

units

Explanation:

D Given,
$x + y = 8$
$x + y = 12$
Latus rectum $= 4 \times$ (dist. between latusrectum and tangent at vertex)
$l = 4 \times | \frac{12 - 8}{\sqrt{1^{2} + 1^{2}}} |$
$l = 4 \times 2 \sqrt{2}$
$l = 8 \sqrt{2}$ \{from (i) \& (ii) $}$

WB JEE-2020

Parabola

120953 The distance between the vertex and the focus to the parabola $x^{2} - 2 x - 3 y - 2 = 0$ is

1

\frac{4}{5}

2

\frac{3}{4}

3

\frac{1}{2}

4

\frac{5}{6}

Explanation:

B Given,
$x^{2} - 2 x - 3 y - 2 = 0$
$x^{2} - 2 x + (1)^{2} - (1)^{2} - 3 y - 2 = 0$
$(x - 1)^{2} = 3 y + 3$
$(x - 1)^{2} = 3 (y + 1)$
We know that, distance between vertex and focus of parabola is a.
From equation (i),
$∴ 4 a = 3$
$a = \frac{3}{4}$

AP EAMCET-2016

Parabola

120954 The locus of the mid-points of all chords of the parabola $y^{2} = 4$ ax through its vertex is another parabola with directrix

1

x = - a

2

x = a

3

x = 0

4

x = - \frac{a}{2}

Explanation:

D Given,
$y^{2} = 4 a x$
Any point on parabola (at $^{2}, 2 a t)$
Chord is passed through vertex $(0, 0)$
Midpoint may be $(\frac{{at}^{2}}{2}$ , at $) = (h, k)$
Where, $h = \frac{{at}^{2}}{2}, k = at$
$2 h = a {(\frac{k}{a})}^{2}$
$2 ah = k^{2}$
Replacing $h$ by $x$ and $k$ by $y$ , we get-
$2 a x = y^{2}$
$y^{2} = 4 (\frac{a}{2}) x$ Thus, directrix is, $x = \frac{- a}{2}$

WB JEE-2016

Parabola

120955 The length of the latus rectum of the parabola $20 (x^{2} + y^{2} - 6 x - 2 y + 10) = (4 x - 2 y - 5)^{2}$ , is

1

\frac{\sqrt{5}}{2}

2

2 \sqrt{5}

3

\sqrt{5}

4

4 \sqrt{5}

Explanation:

C Given,
$20 (x^{2} + y^{2} - 6 x - 2 y + 10) = (4 x - 2 y - 5)^{2}$
$\Rightarrow 20 ((x - 3)^{2} + (y - 1)^{2}) = (4 x - 2 y - 5)^{2}$
$Focus = (3, 1)$
$Directrix : 4 x - 2 y - 5 = 0$
$∴ Latus rectum, l = 2 \times | \frac{3 \times 4 - 2 \times 1 - 5}{\sqrt{(4)^{2} + (- 2)^{2}}} |$
$l = 2 \times | \frac{5}{\sqrt{20}} |$
$l = 2 \times \frac{5}{2 \sqrt{5}}$
$l = \sqrt{5}$

AP EAMCET-22.04.2019

Parabola

120956 The equation of the latus rectum of a parabola is $x + y = 8$ and the equation of the tangent at the vertex is $x + y = 12$ . Then, the length of the latus rectum is

1

4 \sqrt{2}

units

2

2 \sqrt{2}

units

3 8 units

4

8 \sqrt{2}

units

Explanation:

D Given,
$x + y = 8$
$x + y = 12$
Latus rectum $= 4 \times$ (dist. between latusrectum and tangent at vertex)
$l = 4 \times | \frac{12 - 8}{\sqrt{1^{2} + 1^{2}}} |$
$l = 4 \times 2 \sqrt{2}$
$l = 8 \sqrt{2}$ \{from (i) \& (ii) $}$

WB JEE-2020

Parabola

120953 The distance between the vertex and the focus to the parabola $x^{2} - 2 x - 3 y - 2 = 0$ is

1

\frac{4}{5}

2

\frac{3}{4}

3

\frac{1}{2}

4

\frac{5}{6}

Explanation:

B Given,
$x^{2} - 2 x - 3 y - 2 = 0$
$x^{2} - 2 x + (1)^{2} - (1)^{2} - 3 y - 2 = 0$
$(x - 1)^{2} = 3 y + 3$
$(x - 1)^{2} = 3 (y + 1)$
We know that, distance between vertex and focus of parabola is a.
From equation (i),
$∴ 4 a = 3$
$a = \frac{3}{4}$

AP EAMCET-2016

Parabola

120954 The locus of the mid-points of all chords of the parabola $y^{2} = 4$ ax through its vertex is another parabola with directrix

1

x = - a

2

x = a

3

x = 0

4

x = - \frac{a}{2}

Explanation:

D Given,
$y^{2} = 4 a x$
Any point on parabola (at $^{2}, 2 a t)$
Chord is passed through vertex $(0, 0)$
Midpoint may be $(\frac{{at}^{2}}{2}$ , at $) = (h, k)$
Where, $h = \frac{{at}^{2}}{2}, k = at$
$2 h = a {(\frac{k}{a})}^{2}$
$2 ah = k^{2}$
Replacing $h$ by $x$ and $k$ by $y$ , we get-
$2 a x = y^{2}$
$y^{2} = 4 (\frac{a}{2}) x$ Thus, directrix is, $x = \frac{- a}{2}$

WB JEE-2016

Parabola

120955 The length of the latus rectum of the parabola $20 (x^{2} + y^{2} - 6 x - 2 y + 10) = (4 x - 2 y - 5)^{2}$ , is

1

\frac{\sqrt{5}}{2}

2

2 \sqrt{5}

3

\sqrt{5}

4

4 \sqrt{5}

Explanation:

C Given,
$20 (x^{2} + y^{2} - 6 x - 2 y + 10) = (4 x - 2 y - 5)^{2}$
$\Rightarrow 20 ((x - 3)^{2} + (y - 1)^{2}) = (4 x - 2 y - 5)^{2}$
$Focus = (3, 1)$
$Directrix : 4 x - 2 y - 5 = 0$
$∴ Latus rectum, l = 2 \times | \frac{3 \times 4 - 2 \times 1 - 5}{\sqrt{(4)^{2} + (- 2)^{2}}} |$
$l = 2 \times | \frac{5}{\sqrt{20}} |$
$l = 2 \times \frac{5}{2 \sqrt{5}}$
$l = \sqrt{5}$

AP EAMCET-22.04.2019

Parabola

120956 The equation of the latus rectum of a parabola is $x + y = 8$ and the equation of the tangent at the vertex is $x + y = 12$ . Then, the length of the latus rectum is

1

4 \sqrt{2}

units

2

2 \sqrt{2}

units

3 8 units

4

8 \sqrt{2}

units

Explanation:

D Given,
$x + y = 8$
$x + y = 12$
Latus rectum $= 4 \times$ (dist. between latusrectum and tangent at vertex)
$l = 4 \times | \frac{12 - 8}{\sqrt{1^{2} + 1^{2}}} |$
$l = 4 \times 2 \sqrt{2}$
$l = 8 \sqrt{2}$ \{from (i) \& (ii) $}$

WB JEE-2020

Parabola

120953 The distance between the vertex and the focus to the parabola $x^{2} - 2 x - 3 y - 2 = 0$ is

1

\frac{4}{5}

2

\frac{3}{4}

3

\frac{1}{2}

4

\frac{5}{6}

Explanation:

B Given,
$x^{2} - 2 x - 3 y - 2 = 0$
$x^{2} - 2 x + (1)^{2} - (1)^{2} - 3 y - 2 = 0$
$(x - 1)^{2} = 3 y + 3$
$(x - 1)^{2} = 3 (y + 1)$
We know that, distance between vertex and focus of parabola is a.
From equation (i),
$∴ 4 a = 3$
$a = \frac{3}{4}$

AP EAMCET-2016

Parabola

120954 The locus of the mid-points of all chords of the parabola $y^{2} = 4$ ax through its vertex is another parabola with directrix

1

x = - a

2

x = a

3

x = 0

4

x = - \frac{a}{2}

Explanation:

D Given,
$y^{2} = 4 a x$
Any point on parabola (at $^{2}, 2 a t)$
Chord is passed through vertex $(0, 0)$
Midpoint may be $(\frac{{at}^{2}}{2}$ , at $) = (h, k)$
Where, $h = \frac{{at}^{2}}{2}, k = at$
$2 h = a {(\frac{k}{a})}^{2}$
$2 ah = k^{2}$
Replacing $h$ by $x$ and $k$ by $y$ , we get-
$2 a x = y^{2}$
$y^{2} = 4 (\frac{a}{2}) x$ Thus, directrix is, $x = \frac{- a}{2}$

WB JEE-2016

Parabola

120955 The length of the latus rectum of the parabola $20 (x^{2} + y^{2} - 6 x - 2 y + 10) = (4 x - 2 y - 5)^{2}$ , is

1

\frac{\sqrt{5}}{2}

2

2 \sqrt{5}

3

\sqrt{5}

4

4 \sqrt{5}

Explanation:

C Given,
$20 (x^{2} + y^{2} - 6 x - 2 y + 10) = (4 x - 2 y - 5)^{2}$
$\Rightarrow 20 ((x - 3)^{2} + (y - 1)^{2}) = (4 x - 2 y - 5)^{2}$
$Focus = (3, 1)$
$Directrix : 4 x - 2 y - 5 = 0$
$∴ Latus rectum, l = 2 \times | \frac{3 \times 4 - 2 \times 1 - 5}{\sqrt{(4)^{2} + (- 2)^{2}}} |$
$l = 2 \times | \frac{5}{\sqrt{20}} |$
$l = 2 \times \frac{5}{2 \sqrt{5}}$
$l = \sqrt{5}$

AP EAMCET-22.04.2019

Parabola

120956 The equation of the latus rectum of a parabola is $x + y = 8$ and the equation of the tangent at the vertex is $x + y = 12$ . Then, the length of the latus rectum is

1

4 \sqrt{2}

units

2

2 \sqrt{2}

units

3 8 units

4

8 \sqrt{2}

units

Explanation:

D Given,
$x + y = 8$
$x + y = 12$
Latus rectum $= 4 \times$ (dist. between latusrectum and tangent at vertex)
$l = 4 \times | \frac{12 - 8}{\sqrt{1^{2} + 1^{2}}} |$
$l = 4 \times 2 \sqrt{2}$
$l = 8 \sqrt{2}$ \{from (i) \& (ii) $}$

WB JEE-2020