Question

If the angle between the lines represented by the equation ${y^2} + kxy - {x^2}{\tan ^2}A = 0$ be $2A$, then $k = $
A $0$
B $1$
C $2$
D $\tan A$

Answer 1

Hint: First we will compare the given equation ${y^2} + kxy - {x^2}{\tan ^2}A = 0$ with the general equation $a{x^2} + 2hxy + b{y^2} = 0$ and find the value of $a$, $b$ and $h$. Will put all these values in the formula for the angle between two lines and the given angle is $2A$. Then after solving we will get the value of $k$.

Formula Used: $\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$tan2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$

Complete step by step solution: Given, equation is ${y^2} + kxy - {x^2}{\tan ^2}A = 0$
The general equation is $a{x^2} + 2hxy + b{y^2} = 0$
On comparing, we will get
$a = - {\tan ^2}A$, $h = \dfrac{k}{2}$ and $b = 1$
$\alpha = 2A$
We know the angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Putting all the values
$\tan 2A = \left| {\dfrac{{2\sqrt {{{\left( {\dfrac{k}{2}} \right)}^2} - \left( { - {{\tan }^2}A} \right)\left( 1 \right)} }}{{ - {{\tan }^2}A + 1}}} \right|$
$\tan 2A = \left| {\dfrac{{2\sqrt {\dfrac{{{k^2} + 4{{\tan }^2}A}}{4}} }}{{1 - {{\tan }^2}A}}} \right|$
We know $tan2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
$\dfrac{{2\tan A}}{{1 - {{\tan }^2}A}} = \dfrac{{\sqrt {{k^2} + 4{{\tan }^2}A} }}{{1 - {{\tan }^2}A}}$
After solving, we get
$2\tan A = \sqrt {{k^2} + 4{{\tan }^2}A} $
Squaring both the sides
$4{\tan ^2}A = {k^2} + 4{\tan ^2}A$
${k^2} = 0$
$k = 0$
Hence, the angle between the lines is ${0^ \circ }$.
Therefore, correct option is A

Note: Students should solve the question correctly with concentrations to avoid any calculation mistakes. While finding the value of $a$, $b$ and $h$ they should focus on the sign of coefficient whether it is positive or negative.