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Question: If a pair of perpendicular lines represented by \({x^2} + \alpha {y^2} + 2\beta y = {a^2}\) then \(\...

If a pair of perpendicular lines represented by x2+αy2+2βy=a2{x^2} + \alpha {y^2} + 2\beta y = {a^2} then β\beta is
(A)4a (B)a (C)2a (D)3a  \left( A \right)4a \\\ \left( B \right)a \\\ \left( C \right)2a \\\ \left( D \right)3a \\\

Explanation

Solution

Hint:In this question, we use the concept of pair of straight lines. The second degree equation ax2+2hxy+by2+2gx+2fy+c=0a{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, x2+αy2+2βy=a2{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 x2+αy2+2βy=a2{x^2} + \alpha {y^2} + 2\beta y = {a^2} with ax2+2hxy+by2+2gx+2fy+c=0a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 .
a=1,h=0,b=α,g=0,f=β,c=−a2a = 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

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.