# If a pair of perpendicular lines represented by ${x^2} + \alpha {y^2} + 2\beta y = {a^2}$ then $\beta $ is

$

\left( A \right)4a \\

\left( B \right)a \\

\left( C \right)2a \\

\left( D \right)3a \\

$

Answer

Verified

327k+ views

Hint:In this question, we use the concept of pair of straight lines. The second degree equation $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ represents a pair of straight lines if and only if $\left| {\begin{array}{*{20}{c}}

a&h&g \\

h&b&f \\

g&f&c

\end{array}} \right| = 0$ .

Complete step-by-step answer:

Given, ${x^2} + \alpha {y^2} + 2\beta y = {a^2}$

We know the above second degree equation is a pair of straight lines. So, the determinants $\left| {\begin{array}{*{20}{c}}

a&h&g \\

h&b&f \\

g&f&c

\end{array}} \right|$ become 0.

Now, compare the coefficients of equation ${x^2} + \alpha {y^2} + 2\beta y = {a^2}$ with $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ .

$a = 1,h = 0,b = \alpha ,g = 0,f = \beta ,c = - {a^2}$

Now, $\left| {\begin{array}{*{20}{c}}

a&h&g \\

h&b&f \\

g&f&c

\end{array}} \right| = 0$

\[ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&0&0 \\

0&\alpha &\beta \\

0&\beta &{ - {a^2}}

\end{array}} \right| = 0\]

After solving determinant,

\[{\beta ^2} + {a^2}\alpha = 0..............\left( 1 \right)\]

Now, we know a given pair of straight lines are perpendicular to each other. So, we use the formula of the product of slopes in pairs of straight lines.

In pair of straight line product of slopes $ = \dfrac{a}{b}$

So, product of slopes $ = \dfrac{1}{\alpha }$

We also know the product of slopes of two perpendicular lines are $ - 1$ .

$

\Rightarrow - 1 = \dfrac{1}{\alpha } \\

\Rightarrow \alpha = - 1 \\

$

Put the value $\alpha = - 1$ in (1) equation.

\[

{\beta ^2} + {a^2}\left( { - 1} \right) = 0 \\

\Rightarrow {\beta ^2} - {a^2} = 0 \\

\Rightarrow {\beta ^2} = {a^2} \\

\]

Take Square root on both sides

\[ \Rightarrow \beta = \pm a\]

Now in the option value of \[\beta = a\] .

So, the correct option is (B).

Note: Whenever we face such types of problems we use some important points. First we apply the condition of a pair of straight lines and also the product of slopes of two straight lines. So, after some calculation we can get the required answer.

a&h&g \\

h&b&f \\

g&f&c

\end{array}} \right| = 0$ .

Complete step-by-step answer:

Given, ${x^2} + \alpha {y^2} + 2\beta y = {a^2}$

We know the above second degree equation is a pair of straight lines. So, the determinants $\left| {\begin{array}{*{20}{c}}

a&h&g \\

h&b&f \\

g&f&c

\end{array}} \right|$ become 0.

Now, compare the coefficients of equation ${x^2} + \alpha {y^2} + 2\beta y = {a^2}$ with $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ .

$a = 1,h = 0,b = \alpha ,g = 0,f = \beta ,c = - {a^2}$

Now, $\left| {\begin{array}{*{20}{c}}

a&h&g \\

h&b&f \\

g&f&c

\end{array}} \right| = 0$

\[ \Rightarrow \left| {\begin{array}{*{20}{c}}

1&0&0 \\

0&\alpha &\beta \\

0&\beta &{ - {a^2}}

\end{array}} \right| = 0\]

After solving determinant,

\[{\beta ^2} + {a^2}\alpha = 0..............\left( 1 \right)\]

Now, we know a given pair of straight lines are perpendicular to each other. So, we use the formula of the product of slopes in pairs of straight lines.

In pair of straight line product of slopes $ = \dfrac{a}{b}$

So, product of slopes $ = \dfrac{1}{\alpha }$

We also know the product of slopes of two perpendicular lines are $ - 1$ .

$

\Rightarrow - 1 = \dfrac{1}{\alpha } \\

\Rightarrow \alpha = - 1 \\

$

Put the value $\alpha = - 1$ in (1) equation.

\[

{\beta ^2} + {a^2}\left( { - 1} \right) = 0 \\

\Rightarrow {\beta ^2} - {a^2} = 0 \\

\Rightarrow {\beta ^2} = {a^2} \\

\]

Take Square root on both sides

\[ \Rightarrow \beta = \pm a\]

Now in the option value of \[\beta = a\] .

So, the correct option is (B).

Note: Whenever we face such types of problems we use some important points. First we apply the condition of a pair of straight lines and also the product of slopes of two straight lines. So, after some calculation we can get the required answer.

Last updated date: 02nd Jun 2023

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