QBManager

Straight Line

88783 If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is

1

4 x + 3 y = 12 a

2

3 x + 4 y = 7 a

3

7 x + a y = a (7 - a)

4 ay

= (7 - a) (a - x)

Explanation:

(D) : Since equation of family of lines is
$\frac{x}{a} + \frac{y}{b} = 1$
Sum of intercepts $= a + b = 7$ (given)
$\Rightarrow b = 7 - a$
Substitute $b = 7 - a$ in equation (i), we get
$\frac{x}{a} + \frac{y}{7 - a} = 1$
$(7 - a) x + a y = a (7 - a)$
$a y = (7 - a) (a - x)$

COMEDK-2014

Straight Line

88779 If sum of the slopes of the lines given by $x^{2} - 4 p x y + 8 y^{2} = 0$ is three times their product, then $p =$

1 3

2 4

3

\frac{1}{4}

4

\frac{3}{4}

Explanation:

(D) : Given equation of line,
$x^{2} - 4 pxy + 8 y^{2} = 0$
Here $a = 1, 2 h = - 4 p, b = 8$
Let $m_{1}$ and $m_{2}$ be the slope of the lines given by eq. (i)
$m_{1} + m_{2} = - \frac{2 h}{b} = \frac{4 p}{8} = \frac{p}{2}$ and $m_{1} m_{2} = \frac{a}{b} = \frac{1}{8}$
$m_{1} + m_{2} = 3 m_{1} m_{2} \Rightarrow \frac{p}{2} = 3 (\frac{1}{8}) \Rightarrow p = \frac{3}{4}$

MHT CET-2019

Straight Line

88780 The joint equation of bisectors of angles between lines $x = 5$ and $y = 3$ is

1

(x - 5) (y - 3) = 0

2

x^{2} - y^{2} - 10 x + 6 y + 16 = 0

3

x y = 0

4

x y - 5 x - 3 y + 15 = 0

Explanation:

(B) :
We can say both line passes through point $(5, 3)$ and makes angle $45^{\circ}$ and $135^{\circ}$ with $x$ axis
$∴$ Equation of $1^{st}$ line is,
$y - 3 = \tan 45^{\circ} (x - 5)$ i.e.
$y - 3 = x - 5$
$y - x + 2 = 0$
Equation of $2^{nd}$ line is
$y - 3 = \tan 135^{\circ} (x - 5)$ i.e.
$y + x - 8 = 0$
Required joint equation of line is
$(y - x + 2) (y + x - 8) = 0$
$y^{2} - x y + 2 y + x y - x^{2} + 2 x - 8 y + 8 x - 16 = 0$
$∴ - x^{2} + y^{2} + 10 x - 6 y - 16 = 0 \Rightarrow x^{2} - y^{2} - 10 x + 6 y + 16 = 0$

MHT CET-2016

Straight Line

88782 If $4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$ , the family of lines $a x + b y + c = 0$ is concurrent at one or the other of the two points-

1

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

2

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

3

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

4

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

Explanation:

(A) : Given, the equation of line is
$4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$
$= (2 a + b)^{2} - 3 (2 a + b) c + 2 c^{2} = 0$
$\Rightarrow (2 a + b - 2 c) (2 a + b - c) = 0 \Rightarrow c = 2 a + b$
or $2 c = a + \frac{1}{2} b$
The equation of the family of lines is
$a (x + 2) + b (y + 1) = 0$ or $a (x + 1) + b (y + \frac{1}{2}) = 0$
The point of concurrent $(- 2, - 1)$ or $(- 1, - \frac{1}{2})$ .

BITSAT-2013

Straight Line

88784 If the slope of the lines given by ${ax}^{2} + 2 hxy +$ $b y^{2} = 0$ are in the ratio $3 : 1$ , then $h^{2}$ is equal to

1

\frac{ab}{3}

2

\frac{4 a b}{3}

3

\frac{4 a}{3 b}

4 None of these

Explanation:

(B) : Given, equation of line,
Sum of Slope $(m_{1} + m_{2}) = - \frac{2 h}{b}$
$m_{1} m_{2} = \frac{a}{b}$
Given ratios $= 3 : 1$
$m_{1} + 3 m_{1} = - \frac{2 h}{b}$
and $3 m_{1}^{2} = \frac{a}{b}$
From eq $^{n}$ . (i)
$4 m_{1} = - \frac{2 h}{b}$
$m_{1} = - \frac{h}{2 b}$
Substituting value of $m_{1}$ in equation (ii) -
$3 m_{1}^{2} = \frac{a}{b}$
$3 \times {(- \frac{h}{2 b})}^{2} = \frac{a}{b}$
$3 h^{2} = \frac{4 b^{2} a}{b} \Rightarrow 3 h^{2} = 4 a b \Rightarrow h^{2} = \frac{4 a b}{3}$

JCECE-2015

Straight Line

88783 If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is

1

4 x + 3 y = 12 a

2

3 x + 4 y = 7 a

3

7 x + a y = a (7 - a)

4 ay

= (7 - a) (a - x)

Explanation:

(D) : Since equation of family of lines is
$\frac{x}{a} + \frac{y}{b} = 1$
Sum of intercepts $= a + b = 7$ (given)
$\Rightarrow b = 7 - a$
Substitute $b = 7 - a$ in equation (i), we get
$\frac{x}{a} + \frac{y}{7 - a} = 1$
$(7 - a) x + a y = a (7 - a)$
$a y = (7 - a) (a - x)$

COMEDK-2014

Straight Line

88779 If sum of the slopes of the lines given by $x^{2} - 4 p x y + 8 y^{2} = 0$ is three times their product, then $p =$

1 3

2 4

3

\frac{1}{4}

4

\frac{3}{4}

Explanation:

(D) : Given equation of line,
$x^{2} - 4 pxy + 8 y^{2} = 0$
Here $a = 1, 2 h = - 4 p, b = 8$
Let $m_{1}$ and $m_{2}$ be the slope of the lines given by eq. (i)
$m_{1} + m_{2} = - \frac{2 h}{b} = \frac{4 p}{8} = \frac{p}{2}$ and $m_{1} m_{2} = \frac{a}{b} = \frac{1}{8}$
$m_{1} + m_{2} = 3 m_{1} m_{2} \Rightarrow \frac{p}{2} = 3 (\frac{1}{8}) \Rightarrow p = \frac{3}{4}$

MHT CET-2019

Straight Line

88780 The joint equation of bisectors of angles between lines $x = 5$ and $y = 3$ is

1

(x - 5) (y - 3) = 0

2

x^{2} - y^{2} - 10 x + 6 y + 16 = 0

3

x y = 0

4

x y - 5 x - 3 y + 15 = 0

Explanation:

(B) :
We can say both line passes through point $(5, 3)$ and makes angle $45^{\circ}$ and $135^{\circ}$ with $x$ axis
$∴$ Equation of $1^{st}$ line is,
$y - 3 = \tan 45^{\circ} (x - 5)$ i.e.
$y - 3 = x - 5$
$y - x + 2 = 0$
Equation of $2^{nd}$ line is
$y - 3 = \tan 135^{\circ} (x - 5)$ i.e.
$y + x - 8 = 0$
Required joint equation of line is
$(y - x + 2) (y + x - 8) = 0$
$y^{2} - x y + 2 y + x y - x^{2} + 2 x - 8 y + 8 x - 16 = 0$
$∴ - x^{2} + y^{2} + 10 x - 6 y - 16 = 0 \Rightarrow x^{2} - y^{2} - 10 x + 6 y + 16 = 0$

MHT CET-2016

Straight Line

88782 If $4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$ , the family of lines $a x + b y + c = 0$ is concurrent at one or the other of the two points-

1

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

2

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

3

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

4

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

Explanation:

(A) : Given, the equation of line is
$4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$
$= (2 a + b)^{2} - 3 (2 a + b) c + 2 c^{2} = 0$
$\Rightarrow (2 a + b - 2 c) (2 a + b - c) = 0 \Rightarrow c = 2 a + b$
or $2 c = a + \frac{1}{2} b$
The equation of the family of lines is
$a (x + 2) + b (y + 1) = 0$ or $a (x + 1) + b (y + \frac{1}{2}) = 0$
The point of concurrent $(- 2, - 1)$ or $(- 1, - \frac{1}{2})$ .

BITSAT-2013

Straight Line

88784 If the slope of the lines given by ${ax}^{2} + 2 hxy +$ $b y^{2} = 0$ are in the ratio $3 : 1$ , then $h^{2}$ is equal to

1

\frac{ab}{3}

2

\frac{4 a b}{3}

3

\frac{4 a}{3 b}

4 None of these

Explanation:

(B) : Given, equation of line,
Sum of Slope $(m_{1} + m_{2}) = - \frac{2 h}{b}$
$m_{1} m_{2} = \frac{a}{b}$
Given ratios $= 3 : 1$
$m_{1} + 3 m_{1} = - \frac{2 h}{b}$
and $3 m_{1}^{2} = \frac{a}{b}$
From eq $^{n}$ . (i)
$4 m_{1} = - \frac{2 h}{b}$
$m_{1} = - \frac{h}{2 b}$
Substituting value of $m_{1}$ in equation (ii) -
$3 m_{1}^{2} = \frac{a}{b}$
$3 \times {(- \frac{h}{2 b})}^{2} = \frac{a}{b}$
$3 h^{2} = \frac{4 b^{2} a}{b} \Rightarrow 3 h^{2} = 4 a b \Rightarrow h^{2} = \frac{4 a b}{3}$

JCECE-2015

Straight Line

88783 If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is

1

4 x + 3 y = 12 a

2

3 x + 4 y = 7 a

3

7 x + a y = a (7 - a)

4 ay

= (7 - a) (a - x)

Explanation:

(D) : Since equation of family of lines is
$\frac{x}{a} + \frac{y}{b} = 1$
Sum of intercepts $= a + b = 7$ (given)
$\Rightarrow b = 7 - a$
Substitute $b = 7 - a$ in equation (i), we get
$\frac{x}{a} + \frac{y}{7 - a} = 1$
$(7 - a) x + a y = a (7 - a)$
$a y = (7 - a) (a - x)$

COMEDK-2014

Straight Line

88779 If sum of the slopes of the lines given by $x^{2} - 4 p x y + 8 y^{2} = 0$ is three times their product, then $p =$

1 3

2 4

3

\frac{1}{4}

4

\frac{3}{4}

Explanation:

(D) : Given equation of line,
$x^{2} - 4 pxy + 8 y^{2} = 0$
Here $a = 1, 2 h = - 4 p, b = 8$
Let $m_{1}$ and $m_{2}$ be the slope of the lines given by eq. (i)
$m_{1} + m_{2} = - \frac{2 h}{b} = \frac{4 p}{8} = \frac{p}{2}$ and $m_{1} m_{2} = \frac{a}{b} = \frac{1}{8}$
$m_{1} + m_{2} = 3 m_{1} m_{2} \Rightarrow \frac{p}{2} = 3 (\frac{1}{8}) \Rightarrow p = \frac{3}{4}$

MHT CET-2019

Straight Line

88780 The joint equation of bisectors of angles between lines $x = 5$ and $y = 3$ is

1

(x - 5) (y - 3) = 0

2

x^{2} - y^{2} - 10 x + 6 y + 16 = 0

3

x y = 0

4

x y - 5 x - 3 y + 15 = 0

Explanation:

(B) :
We can say both line passes through point $(5, 3)$ and makes angle $45^{\circ}$ and $135^{\circ}$ with $x$ axis
$∴$ Equation of $1^{st}$ line is,
$y - 3 = \tan 45^{\circ} (x - 5)$ i.e.
$y - 3 = x - 5$
$y - x + 2 = 0$
Equation of $2^{nd}$ line is
$y - 3 = \tan 135^{\circ} (x - 5)$ i.e.
$y + x - 8 = 0$
Required joint equation of line is
$(y - x + 2) (y + x - 8) = 0$
$y^{2} - x y + 2 y + x y - x^{2} + 2 x - 8 y + 8 x - 16 = 0$
$∴ - x^{2} + y^{2} + 10 x - 6 y - 16 = 0 \Rightarrow x^{2} - y^{2} - 10 x + 6 y + 16 = 0$

MHT CET-2016

Straight Line

88782 If $4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$ , the family of lines $a x + b y + c = 0$ is concurrent at one or the other of the two points-

1

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

2

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

3

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

4

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

Explanation:

(A) : Given, the equation of line is
$4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$
$= (2 a + b)^{2} - 3 (2 a + b) c + 2 c^{2} = 0$
$\Rightarrow (2 a + b - 2 c) (2 a + b - c) = 0 \Rightarrow c = 2 a + b$
or $2 c = a + \frac{1}{2} b$
The equation of the family of lines is
$a (x + 2) + b (y + 1) = 0$ or $a (x + 1) + b (y + \frac{1}{2}) = 0$
The point of concurrent $(- 2, - 1)$ or $(- 1, - \frac{1}{2})$ .

BITSAT-2013

Straight Line

88784 If the slope of the lines given by ${ax}^{2} + 2 hxy +$ $b y^{2} = 0$ are in the ratio $3 : 1$ , then $h^{2}$ is equal to

1

\frac{ab}{3}

2

\frac{4 a b}{3}

3

\frac{4 a}{3 b}

4 None of these

Explanation:

(B) : Given, equation of line,
Sum of Slope $(m_{1} + m_{2}) = - \frac{2 h}{b}$
$m_{1} m_{2} = \frac{a}{b}$
Given ratios $= 3 : 1$
$m_{1} + 3 m_{1} = - \frac{2 h}{b}$
and $3 m_{1}^{2} = \frac{a}{b}$
From eq $^{n}$ . (i)
$4 m_{1} = - \frac{2 h}{b}$
$m_{1} = - \frac{h}{2 b}$
Substituting value of $m_{1}$ in equation (ii) -
$3 m_{1}^{2} = \frac{a}{b}$
$3 \times {(- \frac{h}{2 b})}^{2} = \frac{a}{b}$
$3 h^{2} = \frac{4 b^{2} a}{b} \Rightarrow 3 h^{2} = 4 a b \Rightarrow h^{2} = \frac{4 a b}{3}$

JCECE-2015

Straight Line

88783 If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is

1

4 x + 3 y = 12 a

2

3 x + 4 y = 7 a

3

7 x + a y = a (7 - a)

4 ay

= (7 - a) (a - x)

Explanation:

(D) : Since equation of family of lines is
$\frac{x}{a} + \frac{y}{b} = 1$
Sum of intercepts $= a + b = 7$ (given)
$\Rightarrow b = 7 - a$
Substitute $b = 7 - a$ in equation (i), we get
$\frac{x}{a} + \frac{y}{7 - a} = 1$
$(7 - a) x + a y = a (7 - a)$
$a y = (7 - a) (a - x)$

COMEDK-2014

Straight Line

88779 If sum of the slopes of the lines given by $x^{2} - 4 p x y + 8 y^{2} = 0$ is three times their product, then $p =$

1 3

2 4

3

\frac{1}{4}

4

\frac{3}{4}

Explanation:

(D) : Given equation of line,
$x^{2} - 4 pxy + 8 y^{2} = 0$
Here $a = 1, 2 h = - 4 p, b = 8$
Let $m_{1}$ and $m_{2}$ be the slope of the lines given by eq. (i)
$m_{1} + m_{2} = - \frac{2 h}{b} = \frac{4 p}{8} = \frac{p}{2}$ and $m_{1} m_{2} = \frac{a}{b} = \frac{1}{8}$
$m_{1} + m_{2} = 3 m_{1} m_{2} \Rightarrow \frac{p}{2} = 3 (\frac{1}{8}) \Rightarrow p = \frac{3}{4}$

MHT CET-2019

Straight Line

88780 The joint equation of bisectors of angles between lines $x = 5$ and $y = 3$ is

1

(x - 5) (y - 3) = 0

2

x^{2} - y^{2} - 10 x + 6 y + 16 = 0

3

x y = 0

4

x y - 5 x - 3 y + 15 = 0

Explanation:

(B) :
We can say both line passes through point $(5, 3)$ and makes angle $45^{\circ}$ and $135^{\circ}$ with $x$ axis
$∴$ Equation of $1^{st}$ line is,
$y - 3 = \tan 45^{\circ} (x - 5)$ i.e.
$y - 3 = x - 5$
$y - x + 2 = 0$
Equation of $2^{nd}$ line is
$y - 3 = \tan 135^{\circ} (x - 5)$ i.e.
$y + x - 8 = 0$
Required joint equation of line is
$(y - x + 2) (y + x - 8) = 0$
$y^{2} - x y + 2 y + x y - x^{2} + 2 x - 8 y + 8 x - 16 = 0$
$∴ - x^{2} + y^{2} + 10 x - 6 y - 16 = 0 \Rightarrow x^{2} - y^{2} - 10 x + 6 y + 16 = 0$

MHT CET-2016

Straight Line

88782 If $4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$ , the family of lines $a x + b y + c = 0$ is concurrent at one or the other of the two points-

1

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

2

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

3

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

4

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

Explanation:

(A) : Given, the equation of line is
$4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$
$= (2 a + b)^{2} - 3 (2 a + b) c + 2 c^{2} = 0$
$\Rightarrow (2 a + b - 2 c) (2 a + b - c) = 0 \Rightarrow c = 2 a + b$
or $2 c = a + \frac{1}{2} b$
The equation of the family of lines is
$a (x + 2) + b (y + 1) = 0$ or $a (x + 1) + b (y + \frac{1}{2}) = 0$
The point of concurrent $(- 2, - 1)$ or $(- 1, - \frac{1}{2})$ .

BITSAT-2013

Straight Line

88784 If the slope of the lines given by ${ax}^{2} + 2 hxy +$ $b y^{2} = 0$ are in the ratio $3 : 1$ , then $h^{2}$ is equal to

1

\frac{ab}{3}

2

\frac{4 a b}{3}

3

\frac{4 a}{3 b}

4 None of these

Explanation:

(B) : Given, equation of line,
Sum of Slope $(m_{1} + m_{2}) = - \frac{2 h}{b}$
$m_{1} m_{2} = \frac{a}{b}$
Given ratios $= 3 : 1$
$m_{1} + 3 m_{1} = - \frac{2 h}{b}$
and $3 m_{1}^{2} = \frac{a}{b}$
From eq $^{n}$ . (i)
$4 m_{1} = - \frac{2 h}{b}$
$m_{1} = - \frac{h}{2 b}$
Substituting value of $m_{1}$ in equation (ii) -
$3 m_{1}^{2} = \frac{a}{b}$
$3 \times {(- \frac{h}{2 b})}^{2} = \frac{a}{b}$
$3 h^{2} = \frac{4 b^{2} a}{b} \Rightarrow 3 h^{2} = 4 a b \Rightarrow h^{2} = \frac{4 a b}{3}$

JCECE-2015

Straight Line

88783 If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is

1

4 x + 3 y = 12 a

2

3 x + 4 y = 7 a

3

7 x + a y = a (7 - a)

4 ay

= (7 - a) (a - x)

Explanation:

(D) : Since equation of family of lines is
$\frac{x}{a} + \frac{y}{b} = 1$
Sum of intercepts $= a + b = 7$ (given)
$\Rightarrow b = 7 - a$
Substitute $b = 7 - a$ in equation (i), we get
$\frac{x}{a} + \frac{y}{7 - a} = 1$
$(7 - a) x + a y = a (7 - a)$
$a y = (7 - a) (a - x)$

COMEDK-2014

Straight Line

88779 If sum of the slopes of the lines given by $x^{2} - 4 p x y + 8 y^{2} = 0$ is three times their product, then $p =$

1 3

2 4

3

\frac{1}{4}

4

\frac{3}{4}

Explanation:

(D) : Given equation of line,
$x^{2} - 4 pxy + 8 y^{2} = 0$
Here $a = 1, 2 h = - 4 p, b = 8$
Let $m_{1}$ and $m_{2}$ be the slope of the lines given by eq. (i)
$m_{1} + m_{2} = - \frac{2 h}{b} = \frac{4 p}{8} = \frac{p}{2}$ and $m_{1} m_{2} = \frac{a}{b} = \frac{1}{8}$
$m_{1} + m_{2} = 3 m_{1} m_{2} \Rightarrow \frac{p}{2} = 3 (\frac{1}{8}) \Rightarrow p = \frac{3}{4}$

MHT CET-2019

Straight Line

88780 The joint equation of bisectors of angles between lines $x = 5$ and $y = 3$ is

1

(x - 5) (y - 3) = 0

2

x^{2} - y^{2} - 10 x + 6 y + 16 = 0

3

x y = 0

4

x y - 5 x - 3 y + 15 = 0

Explanation:

(B) :
We can say both line passes through point $(5, 3)$ and makes angle $45^{\circ}$ and $135^{\circ}$ with $x$ axis
$∴$ Equation of $1^{st}$ line is,
$y - 3 = \tan 45^{\circ} (x - 5)$ i.e.
$y - 3 = x - 5$
$y - x + 2 = 0$
Equation of $2^{nd}$ line is
$y - 3 = \tan 135^{\circ} (x - 5)$ i.e.
$y + x - 8 = 0$
Required joint equation of line is
$(y - x + 2) (y + x - 8) = 0$
$y^{2} - x y + 2 y + x y - x^{2} + 2 x - 8 y + 8 x - 16 = 0$
$∴ - x^{2} + y^{2} + 10 x - 6 y - 16 = 0 \Rightarrow x^{2} - y^{2} - 10 x + 6 y + 16 = 0$

MHT CET-2016

Straight Line

88782 If $4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$ , the family of lines $a x + b y + c = 0$ is concurrent at one or the other of the two points-

1

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

2

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

3

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

4

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

Explanation:

(A) : Given, the equation of line is
$4 a^{2} + b^{2} + 2 c^{2} + 4 a b - 6 a c - 3 b c = 0$
$= (2 a + b)^{2} - 3 (2 a + b) c + 2 c^{2} = 0$
$\Rightarrow (2 a + b - 2 c) (2 a + b - c) = 0 \Rightarrow c = 2 a + b$
or $2 c = a + \frac{1}{2} b$
The equation of the family of lines is
$a (x + 2) + b (y + 1) = 0$ or $a (x + 1) + b (y + \frac{1}{2}) = 0$
The point of concurrent $(- 2, - 1)$ or $(- 1, - \frac{1}{2})$ .

BITSAT-2013

Straight Line

88784 If the slope of the lines given by ${ax}^{2} + 2 hxy +$ $b y^{2} = 0$ are in the ratio $3 : 1$ , then $h^{2}$ is equal to

1

\frac{ab}{3}

2

\frac{4 a b}{3}

3

\frac{4 a}{3 b}

4 None of these

Explanation:

(B) : Given, equation of line,
Sum of Slope $(m_{1} + m_{2}) = - \frac{2 h}{b}$
$m_{1} m_{2} = \frac{a}{b}$
Given ratios $= 3 : 1$
$m_{1} + 3 m_{1} = - \frac{2 h}{b}$
and $3 m_{1}^{2} = \frac{a}{b}$
From eq $^{n}$ . (i)
$4 m_{1} = - \frac{2 h}{b}$
$m_{1} = - \frac{h}{2 b}$
Substituting value of $m_{1}$ in equation (ii) -
$3 m_{1}^{2} = \frac{a}{b}$
$3 \times {(- \frac{h}{2 b})}^{2} = \frac{a}{b}$
$3 h^{2} = \frac{4 b^{2} a}{b} \Rightarrow 3 h^{2} = 4 a b \Rightarrow h^{2} = \frac{4 a b}{3}$

JCECE-2015