QBManager

Parabola

120086 The vertex and the focus of the parabola $2 y^{2} +$ $5 x - 6 y + 1 = 0$ are respectively

1

(\frac{7}{10}, \frac{3}{2}), (\frac{3}{40}, \frac{3}{2})

2

(\frac{- 7}{10}, \frac{3}{2}), (\frac{53}{40}, \frac{3}{2})

3

(\frac{7}{10}, \frac{- 3}{2}), (\frac{7}{10}, \frac{7}{8})

4

(\frac{- 7}{10}, \frac{- 3}{2}), (\frac{7}{10}, \frac{17}{8})

Explanation:

A Given,
$2 y^{2} + 5 x - 6 y + 1 = 0$
$y^{2} - 3 y + \frac{5}{2} x + \frac{1}{2} = 0$
$y^{2} - 3 y + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{2} x + \frac{1}{2} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} x - \frac{9}{4} + \frac{1}{2} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} x - \frac{7}{4} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} (x - \frac{7}{10}) = 0$
${(y - \frac{3}{2})}^{2} = - \frac{5}{2} (x - \frac{7}{10})$
${(y - \frac{3}{2})}^{2} = - 4 \times \frac{5}{8} (x - \frac{7}{10})$
Hence, vertex is $(\frac{7}{10}, \frac{3}{2})$
and focus is $(\frac{3}{40}, \frac{3}{2})$

AP EAMCET-20.04.2019

Parabola

120087 The parabola $x^{2} = 4$ ay makes an intercept of length $\sqrt{40}$ unit on the line $y = 1 + 2 x$ , then a value of $a$ is

1 2

2 -2

3 -1

4 4

Explanation:

B Given,
$x^{2} = 4 a y$ and $y = 1 + 2 x$
On solving equation (i) and (ii), we get -
$x^{2} = 4 a (1 + 2 x)$
$x^{2} = 4 a + 8 a x$
$x^{2} - 8 a x - 4 a = 0$
So, the given line cuts the parabola at two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$x_{1} + x_{2} = 8 a and x_{1}, x_{2} = - 4 a$
$(\sqrt{40})^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}$
$40 = {(x_{1} - x_{2})}^{2} + {(\frac{x_{1}^{2}}{4 a} - \frac{x_{2}^{2}}{4 a})}^{2}$
$40 = [{(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2}] [1 + \frac{{(x_{1} + x_{2})}^{2}}{16 a^{2}}]$
$40 = (8 a)^{2} - 4 (- 4 a) [1 + \frac{(8 a)^{2}}{16 a^{2}}]$
$40 = 8 a^{2} + 16 a [1 + \frac{64 a^{2}}{16 a^{2}}]$
$40 = 5 (8 a^{2} + 16 a)$
$40 = 40 a^{2} + 80 a$
$1 = a^{2} + 2 a$
$a^{2} + 2 a - 1 = 0$
$(a - 1) (a + 2) = 0$
$a = 1, - 2$

AP EAMCET-20.04.2019

Parabola

120088 The co-ordinates of focus of the parabola $5 x^{2} =$ $- 12 y$ are

1

(\frac{3}{5}, 0)

2

(\frac{- 3}{5}, 0)

3

(0, \frac{3}{5})

4

(0, \frac{- 3}{5})

Explanation:

D Given,
$5 x^{2} = - 12 y$
or
$(x - 0)^{2} = \frac{- 12}{5} y$
or
$(x - 0)^{2} = 4 (\frac{- 3}{5}) y$ Hence, focus is - $(0, \frac{- 3}{5})$

AP EAMCET-18.09.2020

Parabola

120089 If a chord of the parabola $y^{2} = 4 x$ passes through its focus and makes an angle $θ$ with the $X$ -axis, then its length is

1

4 \cos^{2} θ

2

4 \sin^{2} θ

3

4 {cosec}^{2} θ

4

4 \sec^{2} θ

Explanation:

C Given,
$y^{2} = 4 x$
Equation of line, $y = mx$
$y = \tan θ . x$
(put in (i))
$\tan^{2} θ \cdot x^{2} = 4 x$
$x^{2} \tan^{2} θ - 4 x = 0$
$x (4 - \tan^{2} θ x) = 0$
$x = \frac{4}{\tan^{2} θ}, 0$
So, $y = \frac{4}{\tan θ}, 0$
Hence, point is- A $(0, 0)$
$B = (\frac{4}{\tan^{2} θ}, \frac{4}{\tan θ})$
Now, distance $A B = \sqrt{{(\frac{4}{\tan^{2} θ})}^{2} + {(\frac{4}{\tan θ})}^{2}}$
$AB = \frac{4}{\tan θ} \sqrt{\frac{1}{\tan θ} - 1}$
$AB = \frac{4}{\tan θ} \times \frac{1}{\sin θ \cos θ}$
$AB = 4 {cosce}^{2} θ$

AP EAMCET-2011

Parabola

120086 The vertex and the focus of the parabola $2 y^{2} +$ $5 x - 6 y + 1 = 0$ are respectively

1

(\frac{7}{10}, \frac{3}{2}), (\frac{3}{40}, \frac{3}{2})

2

(\frac{- 7}{10}, \frac{3}{2}), (\frac{53}{40}, \frac{3}{2})

3

(\frac{7}{10}, \frac{- 3}{2}), (\frac{7}{10}, \frac{7}{8})

4

(\frac{- 7}{10}, \frac{- 3}{2}), (\frac{7}{10}, \frac{17}{8})

Explanation:

A Given,
$2 y^{2} + 5 x - 6 y + 1 = 0$
$y^{2} - 3 y + \frac{5}{2} x + \frac{1}{2} = 0$
$y^{2} - 3 y + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{2} x + \frac{1}{2} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} x - \frac{9}{4} + \frac{1}{2} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} x - \frac{7}{4} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} (x - \frac{7}{10}) = 0$
${(y - \frac{3}{2})}^{2} = - \frac{5}{2} (x - \frac{7}{10})$
${(y - \frac{3}{2})}^{2} = - 4 \times \frac{5}{8} (x - \frac{7}{10})$
Hence, vertex is $(\frac{7}{10}, \frac{3}{2})$
and focus is $(\frac{3}{40}, \frac{3}{2})$

AP EAMCET-20.04.2019

Parabola

120087 The parabola $x^{2} = 4$ ay makes an intercept of length $\sqrt{40}$ unit on the line $y = 1 + 2 x$ , then a value of $a$ is

1 2

2 -2

3 -1

4 4

Explanation:

B Given,
$x^{2} = 4 a y$ and $y = 1 + 2 x$
On solving equation (i) and (ii), we get -
$x^{2} = 4 a (1 + 2 x)$
$x^{2} = 4 a + 8 a x$
$x^{2} - 8 a x - 4 a = 0$
So, the given line cuts the parabola at two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$x_{1} + x_{2} = 8 a and x_{1}, x_{2} = - 4 a$
$(\sqrt{40})^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}$
$40 = {(x_{1} - x_{2})}^{2} + {(\frac{x_{1}^{2}}{4 a} - \frac{x_{2}^{2}}{4 a})}^{2}$
$40 = [{(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2}] [1 + \frac{{(x_{1} + x_{2})}^{2}}{16 a^{2}}]$
$40 = (8 a)^{2} - 4 (- 4 a) [1 + \frac{(8 a)^{2}}{16 a^{2}}]$
$40 = 8 a^{2} + 16 a [1 + \frac{64 a^{2}}{16 a^{2}}]$
$40 = 5 (8 a^{2} + 16 a)$
$40 = 40 a^{2} + 80 a$
$1 = a^{2} + 2 a$
$a^{2} + 2 a - 1 = 0$
$(a - 1) (a + 2) = 0$
$a = 1, - 2$

AP EAMCET-20.04.2019

Parabola

120088 The co-ordinates of focus of the parabola $5 x^{2} =$ $- 12 y$ are

1

(\frac{3}{5}, 0)

2

(\frac{- 3}{5}, 0)

3

(0, \frac{3}{5})

4

(0, \frac{- 3}{5})

Explanation:

D Given,
$5 x^{2} = - 12 y$
or
$(x - 0)^{2} = \frac{- 12}{5} y$
or
$(x - 0)^{2} = 4 (\frac{- 3}{5}) y$ Hence, focus is - $(0, \frac{- 3}{5})$

AP EAMCET-18.09.2020

Parabola

120089 If a chord of the parabola $y^{2} = 4 x$ passes through its focus and makes an angle $θ$ with the $X$ -axis, then its length is

1

4 \cos^{2} θ

2

4 \sin^{2} θ

3

4 {cosec}^{2} θ

4

4 \sec^{2} θ

Explanation:

C Given,
$y^{2} = 4 x$
Equation of line, $y = mx$
$y = \tan θ . x$
(put in (i))
$\tan^{2} θ \cdot x^{2} = 4 x$
$x^{2} \tan^{2} θ - 4 x = 0$
$x (4 - \tan^{2} θ x) = 0$
$x = \frac{4}{\tan^{2} θ}, 0$
So, $y = \frac{4}{\tan θ}, 0$
Hence, point is- A $(0, 0)$
$B = (\frac{4}{\tan^{2} θ}, \frac{4}{\tan θ})$
Now, distance $A B = \sqrt{{(\frac{4}{\tan^{2} θ})}^{2} + {(\frac{4}{\tan θ})}^{2}}$
$AB = \frac{4}{\tan θ} \sqrt{\frac{1}{\tan θ} - 1}$
$AB = \frac{4}{\tan θ} \times \frac{1}{\sin θ \cos θ}$
$AB = 4 {cosce}^{2} θ$

AP EAMCET-2011

Parabola

120086 The vertex and the focus of the parabola $2 y^{2} +$ $5 x - 6 y + 1 = 0$ are respectively

1

(\frac{7}{10}, \frac{3}{2}), (\frac{3}{40}, \frac{3}{2})

2

(\frac{- 7}{10}, \frac{3}{2}), (\frac{53}{40}, \frac{3}{2})

3

(\frac{7}{10}, \frac{- 3}{2}), (\frac{7}{10}, \frac{7}{8})

4

(\frac{- 7}{10}, \frac{- 3}{2}), (\frac{7}{10}, \frac{17}{8})

Explanation:

A Given,
$2 y^{2} + 5 x - 6 y + 1 = 0$
$y^{2} - 3 y + \frac{5}{2} x + \frac{1}{2} = 0$
$y^{2} - 3 y + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{2} x + \frac{1}{2} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} x - \frac{9}{4} + \frac{1}{2} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} x - \frac{7}{4} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} (x - \frac{7}{10}) = 0$
${(y - \frac{3}{2})}^{2} = - \frac{5}{2} (x - \frac{7}{10})$
${(y - \frac{3}{2})}^{2} = - 4 \times \frac{5}{8} (x - \frac{7}{10})$
Hence, vertex is $(\frac{7}{10}, \frac{3}{2})$
and focus is $(\frac{3}{40}, \frac{3}{2})$

AP EAMCET-20.04.2019

Parabola

120087 The parabola $x^{2} = 4$ ay makes an intercept of length $\sqrt{40}$ unit on the line $y = 1 + 2 x$ , then a value of $a$ is

1 2

2 -2

3 -1

4 4

Explanation:

B Given,
$x^{2} = 4 a y$ and $y = 1 + 2 x$
On solving equation (i) and (ii), we get -
$x^{2} = 4 a (1 + 2 x)$
$x^{2} = 4 a + 8 a x$
$x^{2} - 8 a x - 4 a = 0$
So, the given line cuts the parabola at two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$x_{1} + x_{2} = 8 a and x_{1}, x_{2} = - 4 a$
$(\sqrt{40})^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}$
$40 = {(x_{1} - x_{2})}^{2} + {(\frac{x_{1}^{2}}{4 a} - \frac{x_{2}^{2}}{4 a})}^{2}$
$40 = [{(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2}] [1 + \frac{{(x_{1} + x_{2})}^{2}}{16 a^{2}}]$
$40 = (8 a)^{2} - 4 (- 4 a) [1 + \frac{(8 a)^{2}}{16 a^{2}}]$
$40 = 8 a^{2} + 16 a [1 + \frac{64 a^{2}}{16 a^{2}}]$
$40 = 5 (8 a^{2} + 16 a)$
$40 = 40 a^{2} + 80 a$
$1 = a^{2} + 2 a$
$a^{2} + 2 a - 1 = 0$
$(a - 1) (a + 2) = 0$
$a = 1, - 2$

AP EAMCET-20.04.2019

Parabola

120088 The co-ordinates of focus of the parabola $5 x^{2} =$ $- 12 y$ are

1

(\frac{3}{5}, 0)

2

(\frac{- 3}{5}, 0)

3

(0, \frac{3}{5})

4

(0, \frac{- 3}{5})

Explanation:

D Given,
$5 x^{2} = - 12 y$
or
$(x - 0)^{2} = \frac{- 12}{5} y$
or
$(x - 0)^{2} = 4 (\frac{- 3}{5}) y$ Hence, focus is - $(0, \frac{- 3}{5})$

AP EAMCET-18.09.2020

Parabola

120089 If a chord of the parabola $y^{2} = 4 x$ passes through its focus and makes an angle $θ$ with the $X$ -axis, then its length is

1

4 \cos^{2} θ

2

4 \sin^{2} θ

3

4 {cosec}^{2} θ

4

4 \sec^{2} θ

Explanation:

C Given,
$y^{2} = 4 x$
Equation of line, $y = mx$
$y = \tan θ . x$
(put in (i))
$\tan^{2} θ \cdot x^{2} = 4 x$
$x^{2} \tan^{2} θ - 4 x = 0$
$x (4 - \tan^{2} θ x) = 0$
$x = \frac{4}{\tan^{2} θ}, 0$
So, $y = \frac{4}{\tan θ}, 0$
Hence, point is- A $(0, 0)$
$B = (\frac{4}{\tan^{2} θ}, \frac{4}{\tan θ})$
Now, distance $A B = \sqrt{{(\frac{4}{\tan^{2} θ})}^{2} + {(\frac{4}{\tan θ})}^{2}}$
$AB = \frac{4}{\tan θ} \sqrt{\frac{1}{\tan θ} - 1}$
$AB = \frac{4}{\tan θ} \times \frac{1}{\sin θ \cos θ}$
$AB = 4 {cosce}^{2} θ$

AP EAMCET-2011

Parabola

120086 The vertex and the focus of the parabola $2 y^{2} +$ $5 x - 6 y + 1 = 0$ are respectively

1

(\frac{7}{10}, \frac{3}{2}), (\frac{3}{40}, \frac{3}{2})

2

(\frac{- 7}{10}, \frac{3}{2}), (\frac{53}{40}, \frac{3}{2})

3

(\frac{7}{10}, \frac{- 3}{2}), (\frac{7}{10}, \frac{7}{8})

4

(\frac{- 7}{10}, \frac{- 3}{2}), (\frac{7}{10}, \frac{17}{8})

Explanation:

A Given,
$2 y^{2} + 5 x - 6 y + 1 = 0$
$y^{2} - 3 y + \frac{5}{2} x + \frac{1}{2} = 0$
$y^{2} - 3 y + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{2} x + \frac{1}{2} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} x - \frac{9}{4} + \frac{1}{2} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} x - \frac{7}{4} = 0$
${(y - \frac{3}{2})}^{2} + \frac{5}{2} (x - \frac{7}{10}) = 0$
${(y - \frac{3}{2})}^{2} = - \frac{5}{2} (x - \frac{7}{10})$
${(y - \frac{3}{2})}^{2} = - 4 \times \frac{5}{8} (x - \frac{7}{10})$
Hence, vertex is $(\frac{7}{10}, \frac{3}{2})$
and focus is $(\frac{3}{40}, \frac{3}{2})$

AP EAMCET-20.04.2019

Parabola

120087 The parabola $x^{2} = 4$ ay makes an intercept of length $\sqrt{40}$ unit on the line $y = 1 + 2 x$ , then a value of $a$ is

1 2

2 -2

3 -1

4 4

Explanation:

B Given,
$x^{2} = 4 a y$ and $y = 1 + 2 x$
On solving equation (i) and (ii), we get -
$x^{2} = 4 a (1 + 2 x)$
$x^{2} = 4 a + 8 a x$
$x^{2} - 8 a x - 4 a = 0$
So, the given line cuts the parabola at two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$
$x_{1} + x_{2} = 8 a and x_{1}, x_{2} = - 4 a$
$(\sqrt{40})^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}$
$40 = {(x_{1} - x_{2})}^{2} + {(\frac{x_{1}^{2}}{4 a} - \frac{x_{2}^{2}}{4 a})}^{2}$
$40 = [{(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2}] [1 + \frac{{(x_{1} + x_{2})}^{2}}{16 a^{2}}]$
$40 = (8 a)^{2} - 4 (- 4 a) [1 + \frac{(8 a)^{2}}{16 a^{2}}]$
$40 = 8 a^{2} + 16 a [1 + \frac{64 a^{2}}{16 a^{2}}]$
$40 = 5 (8 a^{2} + 16 a)$
$40 = 40 a^{2} + 80 a$
$1 = a^{2} + 2 a$
$a^{2} + 2 a - 1 = 0$
$(a - 1) (a + 2) = 0$
$a = 1, - 2$

AP EAMCET-20.04.2019

Parabola

120088 The co-ordinates of focus of the parabola $5 x^{2} =$ $- 12 y$ are

1

(\frac{3}{5}, 0)

2

(\frac{- 3}{5}, 0)

3

(0, \frac{3}{5})

4

(0, \frac{- 3}{5})

Explanation:

D Given,
$5 x^{2} = - 12 y$
or
$(x - 0)^{2} = \frac{- 12}{5} y$
or
$(x - 0)^{2} = 4 (\frac{- 3}{5}) y$ Hence, focus is - $(0, \frac{- 3}{5})$

AP EAMCET-18.09.2020

Parabola

120089 If a chord of the parabola $y^{2} = 4 x$ passes through its focus and makes an angle $θ$ with the $X$ -axis, then its length is

1

4 \cos^{2} θ

2

4 \sin^{2} θ

3

4 {cosec}^{2} θ

4

4 \sec^{2} θ

Explanation:

C Given,
$y^{2} = 4 x$
Equation of line, $y = mx$
$y = \tan θ . x$
(put in (i))
$\tan^{2} θ \cdot x^{2} = 4 x$
$x^{2} \tan^{2} θ - 4 x = 0$
$x (4 - \tan^{2} θ x) = 0$
$x = \frac{4}{\tan^{2} θ}, 0$
So, $y = \frac{4}{\tan θ}, 0$
Hence, point is- A $(0, 0)$
$B = (\frac{4}{\tan^{2} θ}, \frac{4}{\tan θ})$
Now, distance $A B = \sqrt{{(\frac{4}{\tan^{2} θ})}^{2} + {(\frac{4}{\tan θ})}^{2}}$
$AB = \frac{4}{\tan θ} \sqrt{\frac{1}{\tan θ} - 1}$
$AB = \frac{4}{\tan θ} \times \frac{1}{\sin θ \cos θ}$
$AB = 4 {cosce}^{2} θ$

AP EAMCET-2011

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