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Parabola

120105 A point on the parabola whose focus is $S (1, - 1)$ and whose vertex is $A (1, 1)$ is

1

(3, \frac{1}{2})

2

(1, 2)

3

(2, \frac{1}{2})

4

(2, 2)

Explanation:

A Given,
focus, $S (1, - 1)$ vertex $A (1, 1)$
Equation, $(x - 1)^{2} = 4 \times (- 2) (y - 1)$
$(x - 1)^{2} = - 8 (y - 1)$
$∴$ Point is $(3, \frac{1}{2})$

AP EAMCET-22.04.2018

Parabola

120106 The length of the latus rectum of a parabola whose focal chord PSQ is such that PS = 3 and $Q S = 2$ is

1

\frac{24}{5}

2

\frac{12}{5}

3

\frac{6}{5}

4

\frac{12}{10}

Explanation:

A Given,
$PS = 3 & QS = 2$
By property of parabola
Length of latusrectum $= 2 (\frac{2 SP \times SQ}{SP + SQ})$
$l = \frac{2 \times 2 \times 3 \times 2}{(3 + 2)}$
$l = \frac{24}{5}$

AP EAMCET-22.09.2020

Parabola

120107 Find the equation of the parabola which passes through $(6, - 2)$ , has its vertex at the origin and its axis along the $y$ -axis.

1

y^{2} = 18 x

2

x^{2} = - 18 y

3

y^{2} = - 18 x

4

x^{2} = 18 y

Explanation:

B Given, A $(6, - 2)$ , vertex $(0, 0)$
Equation of parabola whose axis along y-axis is $x^{2} = 4 a y$ or $x^{2} = - 4 a y$
We have,
$4 a \times (- 2) = (6)^{2}$
$a = \frac{- 36}{8} = \frac{- 9}{2}$
Hence, equation is $- x^{2} = - 4 \times \frac{9}{2} y$
$x^{2} = - 18 y$

AP EAMCET-19.08.2021

Parabola

120108 $A B$ is a chord of the parabola $y^{2} = 4 a x$ with vertex at $A$ . $B C$ is drawn perpendicular to $A B$ meeting the axis at $C$ . the projection of $BC$ on the axis of the parabola is

1 2

2

2 a

3

4 a

4

8 a

Explanation:

C Given,
$y^{2} = 4 ax$
original image
Let, $B = [a t^{2}, 2 a t]$
$y = mx$
$2 at = m ({at}^{2})$
$m = \frac{2}{t}$
Slope of $AB = \frac{2}{t}$
$BC ⊥ AB$
Slope of $B C$ . Slope of $A B = - 1$
Slope of $B C \times \frac{2}{t} = - 1$
$∴$ Slope of $BC = \frac{- t}{2}$
Equation for $BC ({at}^{2}, 2 at)$
$y - 2 a t = \frac{- t}{2} (x - a t^{2})$
$y - 2 a t = \frac{- t}{2} (x - a t^{2})$
$- 2 a t = \frac{t}{2} (x - a t^{2}) [∴ y = 0]$
$4 a = x - a t^{2}$
$x = 4 a + a t^{2}$ Distance between CD $= 4 a + a t^{2} - a t^{2} = 4 a$

Jamia Millia Islamia-2009

Parabola

120105 A point on the parabola whose focus is $S (1, - 1)$ and whose vertex is $A (1, 1)$ is

1

(3, \frac{1}{2})

2

(1, 2)

3

(2, \frac{1}{2})

4

(2, 2)

Explanation:

A Given,
focus, $S (1, - 1)$ vertex $A (1, 1)$
Equation, $(x - 1)^{2} = 4 \times (- 2) (y - 1)$
$(x - 1)^{2} = - 8 (y - 1)$
$∴$ Point is $(3, \frac{1}{2})$

AP EAMCET-22.04.2018

Parabola

120106 The length of the latus rectum of a parabola whose focal chord PSQ is such that PS = 3 and $Q S = 2$ is

1

\frac{24}{5}

2

\frac{12}{5}

3

\frac{6}{5}

4

\frac{12}{10}

Explanation:

A Given,
$PS = 3 & QS = 2$
By property of parabola
Length of latusrectum $= 2 (\frac{2 SP \times SQ}{SP + SQ})$
$l = \frac{2 \times 2 \times 3 \times 2}{(3 + 2)}$
$l = \frac{24}{5}$

AP EAMCET-22.09.2020

Parabola

120107 Find the equation of the parabola which passes through $(6, - 2)$ , has its vertex at the origin and its axis along the $y$ -axis.

1

y^{2} = 18 x

2

x^{2} = - 18 y

3

y^{2} = - 18 x

4

x^{2} = 18 y

Explanation:

B Given, A $(6, - 2)$ , vertex $(0, 0)$
Equation of parabola whose axis along y-axis is $x^{2} = 4 a y$ or $x^{2} = - 4 a y$
We have,
$4 a \times (- 2) = (6)^{2}$
$a = \frac{- 36}{8} = \frac{- 9}{2}$
Hence, equation is $- x^{2} = - 4 \times \frac{9}{2} y$
$x^{2} = - 18 y$

AP EAMCET-19.08.2021

Parabola

120108 $A B$ is a chord of the parabola $y^{2} = 4 a x$ with vertex at $A$ . $B C$ is drawn perpendicular to $A B$ meeting the axis at $C$ . the projection of $BC$ on the axis of the parabola is

1 2

2

2 a

3

4 a

4

8 a

Explanation:

C Given,
$y^{2} = 4 ax$
original image
Let, $B = [a t^{2}, 2 a t]$
$y = mx$
$2 at = m ({at}^{2})$
$m = \frac{2}{t}$
Slope of $AB = \frac{2}{t}$
$BC ⊥ AB$
Slope of $B C$ . Slope of $A B = - 1$
Slope of $B C \times \frac{2}{t} = - 1$
$∴$ Slope of $BC = \frac{- t}{2}$
Equation for $BC ({at}^{2}, 2 at)$
$y - 2 a t = \frac{- t}{2} (x - a t^{2})$
$y - 2 a t = \frac{- t}{2} (x - a t^{2})$
$- 2 a t = \frac{t}{2} (x - a t^{2}) [∴ y = 0]$
$4 a = x - a t^{2}$
$x = 4 a + a t^{2}$ Distance between CD $= 4 a + a t^{2} - a t^{2} = 4 a$

Jamia Millia Islamia-2009

Parabola

120105 A point on the parabola whose focus is $S (1, - 1)$ and whose vertex is $A (1, 1)$ is

1

(3, \frac{1}{2})

2

(1, 2)

3

(2, \frac{1}{2})

4

(2, 2)

Explanation:

A Given,
focus, $S (1, - 1)$ vertex $A (1, 1)$
Equation, $(x - 1)^{2} = 4 \times (- 2) (y - 1)$
$(x - 1)^{2} = - 8 (y - 1)$
$∴$ Point is $(3, \frac{1}{2})$

AP EAMCET-22.04.2018

Parabola

120106 The length of the latus rectum of a parabola whose focal chord PSQ is such that PS = 3 and $Q S = 2$ is

1

\frac{24}{5}

2

\frac{12}{5}

3

\frac{6}{5}

4

\frac{12}{10}

Explanation:

A Given,
$PS = 3 & QS = 2$
By property of parabola
Length of latusrectum $= 2 (\frac{2 SP \times SQ}{SP + SQ})$
$l = \frac{2 \times 2 \times 3 \times 2}{(3 + 2)}$
$l = \frac{24}{5}$

AP EAMCET-22.09.2020

Parabola

120107 Find the equation of the parabola which passes through $(6, - 2)$ , has its vertex at the origin and its axis along the $y$ -axis.

1

y^{2} = 18 x

2

x^{2} = - 18 y

3

y^{2} = - 18 x

4

x^{2} = 18 y

Explanation:

B Given, A $(6, - 2)$ , vertex $(0, 0)$
Equation of parabola whose axis along y-axis is $x^{2} = 4 a y$ or $x^{2} = - 4 a y$
We have,
$4 a \times (- 2) = (6)^{2}$
$a = \frac{- 36}{8} = \frac{- 9}{2}$
Hence, equation is $- x^{2} = - 4 \times \frac{9}{2} y$
$x^{2} = - 18 y$

AP EAMCET-19.08.2021

Parabola

120108 $A B$ is a chord of the parabola $y^{2} = 4 a x$ with vertex at $A$ . $B C$ is drawn perpendicular to $A B$ meeting the axis at $C$ . the projection of $BC$ on the axis of the parabola is

1 2

2

2 a

3

4 a

4

8 a

Explanation:

C Given,
$y^{2} = 4 ax$
original image
Let, $B = [a t^{2}, 2 a t]$
$y = mx$
$2 at = m ({at}^{2})$
$m = \frac{2}{t}$
Slope of $AB = \frac{2}{t}$
$BC ⊥ AB$
Slope of $B C$ . Slope of $A B = - 1$
Slope of $B C \times \frac{2}{t} = - 1$
$∴$ Slope of $BC = \frac{- t}{2}$
Equation for $BC ({at}^{2}, 2 at)$
$y - 2 a t = \frac{- t}{2} (x - a t^{2})$
$y - 2 a t = \frac{- t}{2} (x - a t^{2})$
$- 2 a t = \frac{t}{2} (x - a t^{2}) [∴ y = 0]$
$4 a = x - a t^{2}$
$x = 4 a + a t^{2}$ Distance between CD $= 4 a + a t^{2} - a t^{2} = 4 a$

Jamia Millia Islamia-2009

Parabola

120105 A point on the parabola whose focus is $S (1, - 1)$ and whose vertex is $A (1, 1)$ is

1

(3, \frac{1}{2})

2

(1, 2)

3

(2, \frac{1}{2})

4

(2, 2)

Explanation:

A Given,
focus, $S (1, - 1)$ vertex $A (1, 1)$
Equation, $(x - 1)^{2} = 4 \times (- 2) (y - 1)$
$(x - 1)^{2} = - 8 (y - 1)$
$∴$ Point is $(3, \frac{1}{2})$

AP EAMCET-22.04.2018

Parabola

120106 The length of the latus rectum of a parabola whose focal chord PSQ is such that PS = 3 and $Q S = 2$ is

1

\frac{24}{5}

2

\frac{12}{5}

3

\frac{6}{5}

4

\frac{12}{10}

Explanation:

A Given,
$PS = 3 & QS = 2$
By property of parabola
Length of latusrectum $= 2 (\frac{2 SP \times SQ}{SP + SQ})$
$l = \frac{2 \times 2 \times 3 \times 2}{(3 + 2)}$
$l = \frac{24}{5}$

AP EAMCET-22.09.2020

Parabola

120107 Find the equation of the parabola which passes through $(6, - 2)$ , has its vertex at the origin and its axis along the $y$ -axis.

1

y^{2} = 18 x

2

x^{2} = - 18 y

3

y^{2} = - 18 x

4

x^{2} = 18 y

Explanation:

B Given, A $(6, - 2)$ , vertex $(0, 0)$
Equation of parabola whose axis along y-axis is $x^{2} = 4 a y$ or $x^{2} = - 4 a y$
We have,
$4 a \times (- 2) = (6)^{2}$
$a = \frac{- 36}{8} = \frac{- 9}{2}$
Hence, equation is $- x^{2} = - 4 \times \frac{9}{2} y$
$x^{2} = - 18 y$

AP EAMCET-19.08.2021

Parabola

120108 $A B$ is a chord of the parabola $y^{2} = 4 a x$ with vertex at $A$ . $B C$ is drawn perpendicular to $A B$ meeting the axis at $C$ . the projection of $BC$ on the axis of the parabola is

1 2

2

2 a

3

4 a

4

8 a

Explanation:

C Given,
$y^{2} = 4 ax$
original image
Let, $B = [a t^{2}, 2 a t]$
$y = mx$
$2 at = m ({at}^{2})$
$m = \frac{2}{t}$
Slope of $AB = \frac{2}{t}$
$BC ⊥ AB$
Slope of $B C$ . Slope of $A B = - 1$
Slope of $B C \times \frac{2}{t} = - 1$
$∴$ Slope of $BC = \frac{- t}{2}$
Equation for $BC ({at}^{2}, 2 at)$
$y - 2 a t = \frac{- t}{2} (x - a t^{2})$
$y - 2 a t = \frac{- t}{2} (x - a t^{2})$
$- 2 a t = \frac{t}{2} (x - a t^{2}) [∴ y = 0]$
$4 a = x - a t^{2}$
$x = 4 a + a t^{2}$ Distance between CD $= 4 a + a t^{2} - a t^{2} = 4 a$

Jamia Millia Islamia-2009