QBManager

Co-Ordinate system

88389 The measure of the angle between the lines $x^{2}+2 x y \operatorname{cosec} \alpha+y^{2}=0$ is

1 $\pi-\alpha$

2 $\frac{\pi}{2}-\alpha$

3 $\alpha$

4 $\frac{\pi}{2}+\alpha$

nda

Co-Ordinate system

88390 If the angle between the lines given by the equation $x^{2}-3 x y+\lambda y^{2}+3 x-5 y+2=0, \lambda \geq 0$ is
$\boldsymbol{\operatorname { t a n }}^{-1}\left(\frac{1}{3}\right)$, then $\lambda$

1 $\frac{2}{3}, 40$

2 10

3 $1, \frac{2}{5}$

4 2

MHT CET-2020

Co-Ordinate system

88391 If one of the lines given by $k x^{2}+x y-y^{2}=0$ bisects are the angle between the coordinate axes, then values of $k$ are

1 1,2

2 0,2

3 1,3

4

- 2, 2

Explanation:

(B) : Given,
equation of lines is ${kx}^{2} + xy - y^{2} = 0$
$∴ (- 1) m^{2} + m + k = 0$ , is an auxiliary equation.
Since one line is angle bisector of the coordinate axes, we write $m = \pm 1$ .
When, $m = 1$ , from auxiliary equation, we get $k = 0$ .
When, $m = - 1$ , from auxiliary equation, we get $k = 2$ .

MHT CET-2020

Co-Ordinate system

88392 If the acute angle between the lines $x^{2} - 4 x y + y^{2} = 0$ is $\tan^{- 1} (k)$ , then $k =$

1

\sqrt{3}

2

\frac{1}{3}

3

\frac{1}{\sqrt{3}}

4

\frac{1}{6}

Explanation:

(A) :
Acute angle between the lines is given by
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
Here
$a = 1, 2 h = - 4 \Rightarrow h = - 2, b = 1$
$∴ \tan θ = | \frac{2 \sqrt{4 - 1}}{1 + 1} | \Rightarrow \tan θ = \sqrt{3}$
$θ = \tan^{- 1} (\sqrt{3}) \Rightarrow k = \sqrt{3}$

MHT CET-2020

Co-Ordinate system

88393 The angle between the lines $y^{2} \sin^{2} θ - x y \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{2}

4

\frac{π}{3}

Explanation:

(C) :Given,
$y^{2} \sin^{2} θ - xy \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$
$(\sin^{2} θ) y^{2} - (\sin^{2} θ) (xy) - (\sin^{2} θ) x^{2} = 0$
$(\sin^{2} θ) (y^{2} - xy - x^{2}) = 0 \Rightarrow \sin^{2} θ = 0 \Rightarrow θ = \frac{π}{2}$
$(\sin^{2} θ) (y^{2} - x y - x^{2}) = 0$
$\sin θ = 0$ or $(y^{2} - x y - x^{2}) = 0$
$θ = n π$ or $θ = \frac{π}{2}$ as the sum of the coefficient of $x^{2}$ and
$y^{2}$ in the above equation of the pair of straight line is zero i.e. $C_{x} + C_{y} = 0$

MHT CET-2020

Co-Ordinate system

88389 The measure of the angle between the lines $x^{2} + 2 x y cosec α + y^{2} = 0$ is

1

π - α

2

\frac{π}{2} - α

3

α

4

\frac{π}{2} + α

Explanation:

(B) : Given,
$x^{2} + 2 xy cosec α + y^{2} = 0$
Let $θ$ be the angle between the lines,
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
$= | \frac{2 \sqrt{{cosec}^{2} α - 1}}{1 + 1} | = | \frac{2 \sqrt{\cot^{2} α}}{2} |$
$\tan θ = \cot α \Rightarrow \tan θ = \tan (\frac{π}{2} - α)$
$∴ θ = \frac{π}{2} - α$

nda

Co-Ordinate system

88390 If the angle between the lines given by the equation $x^{2} - 3 x y + λ y^{2} + 3 x - 5 y + 2 = 0, λ \geq 0$ is
$\boldsymbol {t a n}^{- 1} (\frac{1}{3})$ , then $λ$

1

\frac{2}{3}, 40

2 10

3

1, \frac{2}{5}

4 2

Explanation:

(D) : Given,
equation of pair of straight line be $x^{2} - 3 x y + λ y^{2} + 3 x - 5 y + 2 = 0$ .
Here $a = 1, b = λ, h = \frac{- 3}{2}$ and angle $θ$ is given by
$\tan^{- 1} (\frac{1}{3}) = \tan^{- 1} (\frac{2 \sqrt{h^{2} - a b}}{a + b})$
$\frac{1}{3} = \frac{2 \sqrt{\frac{9}{4} - λ}}{1 + λ} \Rightarrow (1 + λ)^{2} = {(6 \sqrt{\frac{9}{4} - λ})}^{2}$
$1 + 2 λ + λ^{2} = 36 (\frac{9}{4} - λ) = 81 - 36 λ$
$λ^{2} + 38 λ - 80 = 0 \Rightarrow (λ + 40) (λ - 2) = 0$
$λ = 2, - 40$

MHT CET-2020

Co-Ordinate system

88391 If one of the lines given by $k x^{2} + x y - y^{2} = 0$ bisects are the angle between the coordinate axes, then values of $k$ are

1 1,2

2 0,2

3 1,3

4

- 2, 2

Explanation:

(B) : Given,
equation of lines is ${kx}^{2} + xy - y^{2} = 0$
$∴ (- 1) m^{2} + m + k = 0$ , is an auxiliary equation.
Since one line is angle bisector of the coordinate axes, we write $m = \pm 1$ .
When, $m = 1$ , from auxiliary equation, we get $k = 0$ .
When, $m = - 1$ , from auxiliary equation, we get $k = 2$ .

MHT CET-2020

Co-Ordinate system

88392 If the acute angle between the lines $x^{2} - 4 x y + y^{2} = 0$ is $\tan^{- 1} (k)$ , then $k =$

1

\sqrt{3}

2

\frac{1}{3}

3

\frac{1}{\sqrt{3}}

4

\frac{1}{6}

Explanation:

(A) :
Acute angle between the lines is given by
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
Here
$a = 1, 2 h = - 4 \Rightarrow h = - 2, b = 1$
$∴ \tan θ = | \frac{2 \sqrt{4 - 1}}{1 + 1} | \Rightarrow \tan θ = \sqrt{3}$
$θ = \tan^{- 1} (\sqrt{3}) \Rightarrow k = \sqrt{3}$

MHT CET-2020

Co-Ordinate system

88393 The angle between the lines $y^{2} \sin^{2} θ - x y \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{2}

4

\frac{π}{3}

Explanation:

(C) :Given,
$y^{2} \sin^{2} θ - xy \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$
$(\sin^{2} θ) y^{2} - (\sin^{2} θ) (xy) - (\sin^{2} θ) x^{2} = 0$
$(\sin^{2} θ) (y^{2} - xy - x^{2}) = 0 \Rightarrow \sin^{2} θ = 0 \Rightarrow θ = \frac{π}{2}$
$(\sin^{2} θ) (y^{2} - x y - x^{2}) = 0$
$\sin θ = 0$ or $(y^{2} - x y - x^{2}) = 0$
$θ = n π$ or $θ = \frac{π}{2}$ as the sum of the coefficient of $x^{2}$ and
$y^{2}$ in the above equation of the pair of straight line is zero i.e. $C_{x} + C_{y} = 0$

MHT CET-2020

Co-Ordinate system

88389 The measure of the angle between the lines $x^{2} + 2 x y cosec α + y^{2} = 0$ is

1

π - α

2

\frac{π}{2} - α

3

α

4

\frac{π}{2} + α

Explanation:

(B) : Given,
$x^{2} + 2 xy cosec α + y^{2} = 0$
Let $θ$ be the angle between the lines,
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
$= | \frac{2 \sqrt{{cosec}^{2} α - 1}}{1 + 1} | = | \frac{2 \sqrt{\cot^{2} α}}{2} |$
$\tan θ = \cot α \Rightarrow \tan θ = \tan (\frac{π}{2} - α)$
$∴ θ = \frac{π}{2} - α$

nda

Co-Ordinate system

88390 If the angle between the lines given by the equation $x^{2} - 3 x y + λ y^{2} + 3 x - 5 y + 2 = 0, λ \geq 0$ is
$\boldsymbol {t a n}^{- 1} (\frac{1}{3})$ , then $λ$

1

\frac{2}{3}, 40

2 10

3

1, \frac{2}{5}

4 2

Explanation:

(D) : Given,
equation of pair of straight line be $x^{2} - 3 x y + λ y^{2} + 3 x - 5 y + 2 = 0$ .
Here $a = 1, b = λ, h = \frac{- 3}{2}$ and angle $θ$ is given by
$\tan^{- 1} (\frac{1}{3}) = \tan^{- 1} (\frac{2 \sqrt{h^{2} - a b}}{a + b})$
$\frac{1}{3} = \frac{2 \sqrt{\frac{9}{4} - λ}}{1 + λ} \Rightarrow (1 + λ)^{2} = {(6 \sqrt{\frac{9}{4} - λ})}^{2}$
$1 + 2 λ + λ^{2} = 36 (\frac{9}{4} - λ) = 81 - 36 λ$
$λ^{2} + 38 λ - 80 = 0 \Rightarrow (λ + 40) (λ - 2) = 0$
$λ = 2, - 40$

MHT CET-2020

Co-Ordinate system

88391 If one of the lines given by $k x^{2} + x y - y^{2} = 0$ bisects are the angle between the coordinate axes, then values of $k$ are

1 1,2

2 0,2

3 1,3

4

- 2, 2

Explanation:

(B) : Given,
equation of lines is ${kx}^{2} + xy - y^{2} = 0$
$∴ (- 1) m^{2} + m + k = 0$ , is an auxiliary equation.
Since one line is angle bisector of the coordinate axes, we write $m = \pm 1$ .
When, $m = 1$ , from auxiliary equation, we get $k = 0$ .
When, $m = - 1$ , from auxiliary equation, we get $k = 2$ .

MHT CET-2020

Co-Ordinate system

88392 If the acute angle between the lines $x^{2} - 4 x y + y^{2} = 0$ is $\tan^{- 1} (k)$ , then $k =$

1

\sqrt{3}

2

\frac{1}{3}

3

\frac{1}{\sqrt{3}}

4

\frac{1}{6}

Explanation:

(A) :
Acute angle between the lines is given by
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
Here
$a = 1, 2 h = - 4 \Rightarrow h = - 2, b = 1$
$∴ \tan θ = | \frac{2 \sqrt{4 - 1}}{1 + 1} | \Rightarrow \tan θ = \sqrt{3}$
$θ = \tan^{- 1} (\sqrt{3}) \Rightarrow k = \sqrt{3}$

MHT CET-2020

Co-Ordinate system

88393 The angle between the lines $y^{2} \sin^{2} θ - x y \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{2}

4

\frac{π}{3}

Explanation:

(C) :Given,
$y^{2} \sin^{2} θ - xy \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$
$(\sin^{2} θ) y^{2} - (\sin^{2} θ) (xy) - (\sin^{2} θ) x^{2} = 0$
$(\sin^{2} θ) (y^{2} - xy - x^{2}) = 0 \Rightarrow \sin^{2} θ = 0 \Rightarrow θ = \frac{π}{2}$
$(\sin^{2} θ) (y^{2} - x y - x^{2}) = 0$
$\sin θ = 0$ or $(y^{2} - x y - x^{2}) = 0$
$θ = n π$ or $θ = \frac{π}{2}$ as the sum of the coefficient of $x^{2}$ and
$y^{2}$ in the above equation of the pair of straight line is zero i.e. $C_{x} + C_{y} = 0$

MHT CET-2020

Co-Ordinate system

88389 The measure of the angle between the lines $x^{2} + 2 x y cosec α + y^{2} = 0$ is

1

π - α

2

\frac{π}{2} - α

3

α

4

\frac{π}{2} + α

Explanation:

(B) : Given,
$x^{2} + 2 xy cosec α + y^{2} = 0$
Let $θ$ be the angle between the lines,
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
$= | \frac{2 \sqrt{{cosec}^{2} α - 1}}{1 + 1} | = | \frac{2 \sqrt{\cot^{2} α}}{2} |$
$\tan θ = \cot α \Rightarrow \tan θ = \tan (\frac{π}{2} - α)$
$∴ θ = \frac{π}{2} - α$

nda

Co-Ordinate system

88390 If the angle between the lines given by the equation $x^{2} - 3 x y + λ y^{2} + 3 x - 5 y + 2 = 0, λ \geq 0$ is
$\boldsymbol {t a n}^{- 1} (\frac{1}{3})$ , then $λ$

1

\frac{2}{3}, 40

2 10

3

1, \frac{2}{5}

4 2

Explanation:

(D) : Given,
equation of pair of straight line be $x^{2} - 3 x y + λ y^{2} + 3 x - 5 y + 2 = 0$ .
Here $a = 1, b = λ, h = \frac{- 3}{2}$ and angle $θ$ is given by
$\tan^{- 1} (\frac{1}{3}) = \tan^{- 1} (\frac{2 \sqrt{h^{2} - a b}}{a + b})$
$\frac{1}{3} = \frac{2 \sqrt{\frac{9}{4} - λ}}{1 + λ} \Rightarrow (1 + λ)^{2} = {(6 \sqrt{\frac{9}{4} - λ})}^{2}$
$1 + 2 λ + λ^{2} = 36 (\frac{9}{4} - λ) = 81 - 36 λ$
$λ^{2} + 38 λ - 80 = 0 \Rightarrow (λ + 40) (λ - 2) = 0$
$λ = 2, - 40$

MHT CET-2020

Co-Ordinate system

88391 If one of the lines given by $k x^{2} + x y - y^{2} = 0$ bisects are the angle between the coordinate axes, then values of $k$ are

1 1,2

2 0,2

3 1,3

4

- 2, 2

Explanation:

(B) : Given,
equation of lines is ${kx}^{2} + xy - y^{2} = 0$
$∴ (- 1) m^{2} + m + k = 0$ , is an auxiliary equation.
Since one line is angle bisector of the coordinate axes, we write $m = \pm 1$ .
When, $m = 1$ , from auxiliary equation, we get $k = 0$ .
When, $m = - 1$ , from auxiliary equation, we get $k = 2$ .

MHT CET-2020

Co-Ordinate system

88392 If the acute angle between the lines $x^{2} - 4 x y + y^{2} = 0$ is $\tan^{- 1} (k)$ , then $k =$

1

\sqrt{3}

2

\frac{1}{3}

3

\frac{1}{\sqrt{3}}

4

\frac{1}{6}

Explanation:

(A) :
Acute angle between the lines is given by
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
Here
$a = 1, 2 h = - 4 \Rightarrow h = - 2, b = 1$
$∴ \tan θ = | \frac{2 \sqrt{4 - 1}}{1 + 1} | \Rightarrow \tan θ = \sqrt{3}$
$θ = \tan^{- 1} (\sqrt{3}) \Rightarrow k = \sqrt{3}$

MHT CET-2020

Co-Ordinate system

88393 The angle between the lines $y^{2} \sin^{2} θ - x y \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{2}

4

\frac{π}{3}

Explanation:

(C) :Given,
$y^{2} \sin^{2} θ - xy \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$
$(\sin^{2} θ) y^{2} - (\sin^{2} θ) (xy) - (\sin^{2} θ) x^{2} = 0$
$(\sin^{2} θ) (y^{2} - xy - x^{2}) = 0 \Rightarrow \sin^{2} θ = 0 \Rightarrow θ = \frac{π}{2}$
$(\sin^{2} θ) (y^{2} - x y - x^{2}) = 0$
$\sin θ = 0$ or $(y^{2} - x y - x^{2}) = 0$
$θ = n π$ or $θ = \frac{π}{2}$ as the sum of the coefficient of $x^{2}$ and
$y^{2}$ in the above equation of the pair of straight line is zero i.e. $C_{x} + C_{y} = 0$

MHT CET-2020

Co-Ordinate system

88389 The measure of the angle between the lines $x^{2} + 2 x y cosec α + y^{2} = 0$ is

1

π - α

2

\frac{π}{2} - α

3

α

4

\frac{π}{2} + α

Explanation:

(B) : Given,
$x^{2} + 2 xy cosec α + y^{2} = 0$
Let $θ$ be the angle between the lines,
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
$= | \frac{2 \sqrt{{cosec}^{2} α - 1}}{1 + 1} | = | \frac{2 \sqrt{\cot^{2} α}}{2} |$
$\tan θ = \cot α \Rightarrow \tan θ = \tan (\frac{π}{2} - α)$
$∴ θ = \frac{π}{2} - α$

nda

Co-Ordinate system

88390 If the angle between the lines given by the equation $x^{2} - 3 x y + λ y^{2} + 3 x - 5 y + 2 = 0, λ \geq 0$ is
$\boldsymbol {t a n}^{- 1} (\frac{1}{3})$ , then $λ$

1

\frac{2}{3}, 40

2 10

3

1, \frac{2}{5}

4 2

Explanation:

(D) : Given,
equation of pair of straight line be $x^{2} - 3 x y + λ y^{2} + 3 x - 5 y + 2 = 0$ .
Here $a = 1, b = λ, h = \frac{- 3}{2}$ and angle $θ$ is given by
$\tan^{- 1} (\frac{1}{3}) = \tan^{- 1} (\frac{2 \sqrt{h^{2} - a b}}{a + b})$
$\frac{1}{3} = \frac{2 \sqrt{\frac{9}{4} - λ}}{1 + λ} \Rightarrow (1 + λ)^{2} = {(6 \sqrt{\frac{9}{4} - λ})}^{2}$
$1 + 2 λ + λ^{2} = 36 (\frac{9}{4} - λ) = 81 - 36 λ$
$λ^{2} + 38 λ - 80 = 0 \Rightarrow (λ + 40) (λ - 2) = 0$
$λ = 2, - 40$

MHT CET-2020

Co-Ordinate system

88391 If one of the lines given by $k x^{2} + x y - y^{2} = 0$ bisects are the angle between the coordinate axes, then values of $k$ are

1 1,2

2 0,2

3 1,3

4

- 2, 2

Explanation:

(B) : Given,
equation of lines is ${kx}^{2} + xy - y^{2} = 0$
$∴ (- 1) m^{2} + m + k = 0$ , is an auxiliary equation.
Since one line is angle bisector of the coordinate axes, we write $m = \pm 1$ .
When, $m = 1$ , from auxiliary equation, we get $k = 0$ .
When, $m = - 1$ , from auxiliary equation, we get $k = 2$ .

MHT CET-2020

Co-Ordinate system

88392 If the acute angle between the lines $x^{2} - 4 x y + y^{2} = 0$ is $\tan^{- 1} (k)$ , then $k =$

1

\sqrt{3}

2

\frac{1}{3}

3

\frac{1}{\sqrt{3}}

4

\frac{1}{6}

Explanation:

(A) :
Acute angle between the lines is given by
$\tan θ = | \frac{2 \sqrt{h^{2} - ab}}{a + b} |$
Here
$a = 1, 2 h = - 4 \Rightarrow h = - 2, b = 1$
$∴ \tan θ = | \frac{2 \sqrt{4 - 1}}{1 + 1} | \Rightarrow \tan θ = \sqrt{3}$
$θ = \tan^{- 1} (\sqrt{3}) \Rightarrow k = \sqrt{3}$

MHT CET-2020

Co-Ordinate system

88393 The angle between the lines $y^{2} \sin^{2} θ - x y \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$ is

1

\frac{π}{6}

2

\frac{π}{4}

3

\frac{π}{2}

4

\frac{π}{3}

Explanation:

(C) :Given,
$y^{2} \sin^{2} θ - xy \sin^{2} θ + x^{2} (\cos^{2} θ - 1) = 0$
$(\sin^{2} θ) y^{2} - (\sin^{2} θ) (xy) - (\sin^{2} θ) x^{2} = 0$
$(\sin^{2} θ) (y^{2} - xy - x^{2}) = 0 \Rightarrow \sin^{2} θ = 0 \Rightarrow θ = \frac{π}{2}$
$(\sin^{2} θ) (y^{2} - x y - x^{2}) = 0$
$\sin θ = 0$ or $(y^{2} - x y - x^{2}) = 0$
$θ = n π$ or $θ = \frac{π}{2}$ as the sum of the coefficient of $x^{2}$ and
$y^{2}$ in the above equation of the pair of straight line is zero i.e. $C_{x} + C_{y} = 0$

MHT CET-2020

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