QBManager

Hyperbola

120832 The product of distances from any point on the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ to its two asymptotes is

1

\frac{25}{144}

2

\frac{144}{25}

3

\frac{12}{5}

4 None of these

Explanation:

B Given equation is hyperbola,
$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
On comparing standerd hyperbola equation,
$a^{2} = 16 and b^{2} = 9$
Where,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$d = \frac{a^{2} b^{2}}{a^{2} + b^{2}} = \frac{16 \times 9}{16 + 9}$
$d = \frac{144}{25}$

EAMCET-1993

Hyperbola

120833 If the equation of one asymptote of the hyperbola $14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 91 = 0$ is $7 x + 5 y - 3 = 0$ , then the other asymptote is

1

2 x - 4 y + 1 = 0

2

2 x + 4 y + 1 = 0

3

2 x - 4 y - 1 = 0

4

2 x + 4 y - 1 = 0

Explanation:

B Given,
$14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 91 = 0$
We know that, euqation of the pair of line -
$Δ = a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
Here, $a = 14, b = 20, f = 19, g = \frac{1}{2}, h = \frac{- 7}{2}$
$14 (20) c + 2 (19) (\frac{1}{2}) (\frac{- 7}{2}) - 14 {(\frac{7}{2})}^{2} - 20 {(\frac{1}{2})}^{2} - c (19)^{2} = 0$
$81 c = - 243$
$c = - 3$
$∴ 14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 3 = (7 x + 5 y - 3) (2 x +$ $4 y + 1)$ .
So, other asymptote is :
$2 x + 4 y + 1 = 0$

AP EAMCET-24.04.2018

Hyperbola

120834 The asymptotes of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , with any tangent to the hyperbola form a triangle whose area is $a^{2} \tan (α)$ . Then its eccentricity equals

1

\sec (α)

2

cosec (α)

3

\sec^{2} (α)

4

{cosec}^{2} (α)

Explanation:

A : Given,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Any tangent to the hyperbola from a triangle with the assymptotes which have always constant area and that is equal to $(a b)$ .
$a^{2} \tan (α) = ab$
$\tan α = \frac{b}{a}$
$e^{2} = 1 + \frac{b^{2}}{a^{2}} [e >> 1]$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + (\tan α)^{2}}$
$e = \sqrt{\sec^{2} α}$
$e = \sec α$

AP EAMCET-19.08.2021

Hyperbola

120835 The equation of the hyperbola whose asymptotes are the lines $3 x + 4 y - 2 = 0$ , $2 x + y + 1 = 0$ and which passes through the point $(1, 1)$ is

1

6 x^{2} + 11 x y + 4 y^{2} - 30 x + 2 y + 7 = 0

2

6 x^{2} + 11 x y + 4 y^{2} - x + 2 y - 22 = 0

3

6 x^{2} + 11 x y + 4 y^{2} - x + 2 y + 22 = 0

4

6 x^{2} + 11 x y + 4 y^{2} - 3 x - 7 y - 11 = 0

Explanation:

B Given,
$l_{1} : 3 x + 4 y - 2 = 0$
$l_{2} : 2 x + y + 1 + 0$
Equation of hyperbola,
$(3 x + 4 y - 2) (2 x + y + 1) + k = 0$
On passing at point $(1, 1)$ we get -
$(3 + 4 - 2) (2 + 1 + 1) + k = 0$
$20 + k = 0$
Since, it is passes through the point $(1, 1)$
Then,
$(3 \times 1 + 4 \times 1 - 2) (2 \times 1 + 1 + 1) + k = 0$
$5 \times 4 + k = 0$
$20 + k = 0$
$k = - 20$
Hence, equation of hyperbola is -
$6 x^{2} + 11 x y + 4 y^{2} - x + 2 y - 22 = 0$

AP EAMCET-22.04.2018

Hyperbola

120832 The product of distances from any point on the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ to its two asymptotes is

1

\frac{25}{144}

2

\frac{144}{25}

3

\frac{12}{5}

4 None of these

Explanation:

B Given equation is hyperbola,
$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
On comparing standerd hyperbola equation,
$a^{2} = 16 and b^{2} = 9$
Where,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$d = \frac{a^{2} b^{2}}{a^{2} + b^{2}} = \frac{16 \times 9}{16 + 9}$
$d = \frac{144}{25}$

EAMCET-1993

Hyperbola

120833 If the equation of one asymptote of the hyperbola $14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 91 = 0$ is $7 x + 5 y - 3 = 0$ , then the other asymptote is

1

2 x - 4 y + 1 = 0

2

2 x + 4 y + 1 = 0

3

2 x - 4 y - 1 = 0

4

2 x + 4 y - 1 = 0

Explanation:

B Given,
$14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 91 = 0$
We know that, euqation of the pair of line -
$Δ = a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
Here, $a = 14, b = 20, f = 19, g = \frac{1}{2}, h = \frac{- 7}{2}$
$14 (20) c + 2 (19) (\frac{1}{2}) (\frac{- 7}{2}) - 14 {(\frac{7}{2})}^{2} - 20 {(\frac{1}{2})}^{2} - c (19)^{2} = 0$
$81 c = - 243$
$c = - 3$
$∴ 14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 3 = (7 x + 5 y - 3) (2 x +$ $4 y + 1)$ .
So, other asymptote is :
$2 x + 4 y + 1 = 0$

AP EAMCET-24.04.2018

Hyperbola

120834 The asymptotes of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , with any tangent to the hyperbola form a triangle whose area is $a^{2} \tan (α)$ . Then its eccentricity equals

1

\sec (α)

2

cosec (α)

3

\sec^{2} (α)

4

{cosec}^{2} (α)

Explanation:

A : Given,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Any tangent to the hyperbola from a triangle with the assymptotes which have always constant area and that is equal to $(a b)$ .
$a^{2} \tan (α) = ab$
$\tan α = \frac{b}{a}$
$e^{2} = 1 + \frac{b^{2}}{a^{2}} [e >> 1]$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + (\tan α)^{2}}$
$e = \sqrt{\sec^{2} α}$
$e = \sec α$

AP EAMCET-19.08.2021

Hyperbola

120835 The equation of the hyperbola whose asymptotes are the lines $3 x + 4 y - 2 = 0$ , $2 x + y + 1 = 0$ and which passes through the point $(1, 1)$ is

1

6 x^{2} + 11 x y + 4 y^{2} - 30 x + 2 y + 7 = 0

2

6 x^{2} + 11 x y + 4 y^{2} - x + 2 y - 22 = 0

3

6 x^{2} + 11 x y + 4 y^{2} - x + 2 y + 22 = 0

4

6 x^{2} + 11 x y + 4 y^{2} - 3 x - 7 y - 11 = 0

Explanation:

B Given,
$l_{1} : 3 x + 4 y - 2 = 0$
$l_{2} : 2 x + y + 1 + 0$
Equation of hyperbola,
$(3 x + 4 y - 2) (2 x + y + 1) + k = 0$
On passing at point $(1, 1)$ we get -
$(3 + 4 - 2) (2 + 1 + 1) + k = 0$
$20 + k = 0$
Since, it is passes through the point $(1, 1)$
Then,
$(3 \times 1 + 4 \times 1 - 2) (2 \times 1 + 1 + 1) + k = 0$
$5 \times 4 + k = 0$
$20 + k = 0$
$k = - 20$
Hence, equation of hyperbola is -
$6 x^{2} + 11 x y + 4 y^{2} - x + 2 y - 22 = 0$

AP EAMCET-22.04.2018

Hyperbola

120832 The product of distances from any point on the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ to its two asymptotes is

1

\frac{25}{144}

2

\frac{144}{25}

3

\frac{12}{5}

4 None of these

Explanation:

B Given equation is hyperbola,
$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
On comparing standerd hyperbola equation,
$a^{2} = 16 and b^{2} = 9$
Where,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$d = \frac{a^{2} b^{2}}{a^{2} + b^{2}} = \frac{16 \times 9}{16 + 9}$
$d = \frac{144}{25}$

EAMCET-1993

Hyperbola

120833 If the equation of one asymptote of the hyperbola $14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 91 = 0$ is $7 x + 5 y - 3 = 0$ , then the other asymptote is

1

2 x - 4 y + 1 = 0

2

2 x + 4 y + 1 = 0

3

2 x - 4 y - 1 = 0

4

2 x + 4 y - 1 = 0

Explanation:

B Given,
$14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 91 = 0$
We know that, euqation of the pair of line -
$Δ = a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
Here, $a = 14, b = 20, f = 19, g = \frac{1}{2}, h = \frac{- 7}{2}$
$14 (20) c + 2 (19) (\frac{1}{2}) (\frac{- 7}{2}) - 14 {(\frac{7}{2})}^{2} - 20 {(\frac{1}{2})}^{2} - c (19)^{2} = 0$
$81 c = - 243$
$c = - 3$
$∴ 14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 3 = (7 x + 5 y - 3) (2 x +$ $4 y + 1)$ .
So, other asymptote is :
$2 x + 4 y + 1 = 0$

AP EAMCET-24.04.2018

Hyperbola

120834 The asymptotes of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , with any tangent to the hyperbola form a triangle whose area is $a^{2} \tan (α)$ . Then its eccentricity equals

1

\sec (α)

2

cosec (α)

3

\sec^{2} (α)

4

{cosec}^{2} (α)

Explanation:

A : Given,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Any tangent to the hyperbola from a triangle with the assymptotes which have always constant area and that is equal to $(a b)$ .
$a^{2} \tan (α) = ab$
$\tan α = \frac{b}{a}$
$e^{2} = 1 + \frac{b^{2}}{a^{2}} [e >> 1]$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + (\tan α)^{2}}$
$e = \sqrt{\sec^{2} α}$
$e = \sec α$

AP EAMCET-19.08.2021

Hyperbola

120835 The equation of the hyperbola whose asymptotes are the lines $3 x + 4 y - 2 = 0$ , $2 x + y + 1 = 0$ and which passes through the point $(1, 1)$ is

1

6 x^{2} + 11 x y + 4 y^{2} - 30 x + 2 y + 7 = 0

2

6 x^{2} + 11 x y + 4 y^{2} - x + 2 y - 22 = 0

3

6 x^{2} + 11 x y + 4 y^{2} - x + 2 y + 22 = 0

4

6 x^{2} + 11 x y + 4 y^{2} - 3 x - 7 y - 11 = 0

Explanation:

B Given,
$l_{1} : 3 x + 4 y - 2 = 0$
$l_{2} : 2 x + y + 1 + 0$
Equation of hyperbola,
$(3 x + 4 y - 2) (2 x + y + 1) + k = 0$
On passing at point $(1, 1)$ we get -
$(3 + 4 - 2) (2 + 1 + 1) + k = 0$
$20 + k = 0$
Since, it is passes through the point $(1, 1)$
Then,
$(3 \times 1 + 4 \times 1 - 2) (2 \times 1 + 1 + 1) + k = 0$
$5 \times 4 + k = 0$
$20 + k = 0$
$k = - 20$
Hence, equation of hyperbola is -
$6 x^{2} + 11 x y + 4 y^{2} - x + 2 y - 22 = 0$

AP EAMCET-22.04.2018

Hyperbola

120832 The product of distances from any point on the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ to its two asymptotes is

1

\frac{25}{144}

2

\frac{144}{25}

3

\frac{12}{5}

4 None of these

Explanation:

B Given equation is hyperbola,
$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
On comparing standerd hyperbola equation,
$a^{2} = 16 and b^{2} = 9$
Where,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$d = \frac{a^{2} b^{2}}{a^{2} + b^{2}} = \frac{16 \times 9}{16 + 9}$
$d = \frac{144}{25}$

EAMCET-1993

Hyperbola

120833 If the equation of one asymptote of the hyperbola $14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 91 = 0$ is $7 x + 5 y - 3 = 0$ , then the other asymptote is

1

2 x - 4 y + 1 = 0

2

2 x + 4 y + 1 = 0

3

2 x - 4 y - 1 = 0

4

2 x + 4 y - 1 = 0

Explanation:

B Given,
$14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 91 = 0$
We know that, euqation of the pair of line -
$Δ = a b c + 2 f g h - a f^{2} - b^{2} - c h^{2} = 0$
Here, $a = 14, b = 20, f = 19, g = \frac{1}{2}, h = \frac{- 7}{2}$
$14 (20) c + 2 (19) (\frac{1}{2}) (\frac{- 7}{2}) - 14 {(\frac{7}{2})}^{2} - 20 {(\frac{1}{2})}^{2} - c (19)^{2} = 0$
$81 c = - 243$
$c = - 3$
$∴ 14 x^{2} + 38 x y + 20 y^{2} + x - 7 y - 3 = (7 x + 5 y - 3) (2 x +$ $4 y + 1)$ .
So, other asymptote is :
$2 x + 4 y + 1 = 0$

AP EAMCET-24.04.2018

Hyperbola

120834 The asymptotes of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , with any tangent to the hyperbola form a triangle whose area is $a^{2} \tan (α)$ . Then its eccentricity equals

1

\sec (α)

2

cosec (α)

3

\sec^{2} (α)

4

{cosec}^{2} (α)

Explanation:

A : Given,
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Any tangent to the hyperbola from a triangle with the assymptotes which have always constant area and that is equal to $(a b)$ .
$a^{2} \tan (α) = ab$
$\tan α = \frac{b}{a}$
$e^{2} = 1 + \frac{b^{2}}{a^{2}} [e >> 1]$
$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$
$e = \sqrt{1 + (\tan α)^{2}}$
$e = \sqrt{\sec^{2} α}$
$e = \sec α$

AP EAMCET-19.08.2021

Hyperbola

120835 The equation of the hyperbola whose asymptotes are the lines $3 x + 4 y - 2 = 0$ , $2 x + y + 1 = 0$ and which passes through the point $(1, 1)$ is

1

6 x^{2} + 11 x y + 4 y^{2} - 30 x + 2 y + 7 = 0

2

6 x^{2} + 11 x y + 4 y^{2} - x + 2 y - 22 = 0

3

6 x^{2} + 11 x y + 4 y^{2} - x + 2 y + 22 = 0

4

6 x^{2} + 11 x y + 4 y^{2} - 3 x - 7 y - 11 = 0

Explanation:

B Given,
$l_{1} : 3 x + 4 y - 2 = 0$
$l_{2} : 2 x + y + 1 + 0$
Equation of hyperbola,
$(3 x + 4 y - 2) (2 x + y + 1) + k = 0$
On passing at point $(1, 1)$ we get -
$(3 + 4 - 2) (2 + 1 + 1) + k = 0$
$20 + k = 0$
Since, it is passes through the point $(1, 1)$
Then,
$(3 \times 1 + 4 \times 1 - 2) (2 \times 1 + 1 + 1) + k = 0$
$5 \times 4 + k = 0$
$20 + k = 0$
$k = - 20$
Hence, equation of hyperbola is -
$6 x^{2} + 11 x y + 4 y^{2} - x + 2 y - 22 = 0$

AP EAMCET-22.04.2018

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