
Angle between the line joining the origin to the point of intersection of the curves $2{x^2} + 3{y^2} + 10x = 0$ and $3{x^2} + 5{y^2} + 16x = 0$ is
A. ${\tan ^{ - 1}}\left( {\dfrac{3}{2}} \right)$
B. ${\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right)$
C. ${90^ \circ }$
D. None of these
Answer
163.5k+ views
Hint: We have to find the angle between the line joining the origin to the point of intersection of the given curves. First, we have to find the equation of the line. Then compare the resultant equation with the general to get the value of $a$, $b$ and $h$. Then put these values in formula to get the angle.
Formula Used: $\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Complete step by step solution: Given, equations of curve $2{x^2} + 3{y^2} + 10x = 0$ and $3{x^2} + 5{y^2} + 16x = 0$
The equation of the curve through the point of intersection of the given curves
$2{x^2} + 3{y^2} + 10x + \lambda (3{x^2} + 5{y^2} + 16x) = 0$ ….(1)
We know If the equation of the line through the origin, then the equation must be homogeneous equation of second degree
equation must be of type ${x^2} + {y^2} = 0$
Therefore, the coefficient of $x$ is zero.
So, $10 + 16\lambda = 0$
$16\lambda = - 10$
$\lambda = \dfrac{{ - 5}}{8}$
Putting the value of $\lambda $ in equation (1)
$2{x^2} + 3{y^2} - \dfrac{5}{8}(3{x^2} + 5{y^2}) = 0$
$16{x^2} + 24{y^2} - 15{x^2} - 25{y^2} = 0$
${x^2} - {y^2} = 0$
The general equation is $a{x^2} + 2hxy + b{y^2} = 0$
On comparing, we will get
$a = 1$, $h = 0$ and $b = - 1$
We know the formula for angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$\tan \alpha = \left| {\dfrac{{2\sqrt {0 - (1)( - 1)} }}{{1 + ( - 1)}}} \right|$
After solving, we get
$\tan \alpha = \dfrac{2}{0}$
Taking ${\tan ^{ - 1}}$ on both side.
${\tan ^{ - 1}}\left( {\tan \alpha } \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{0}} \right)$
$\alpha = {90^ \circ }$
Hence, the angle between the lines is ${90^ \circ }$
Therefore, option C is correct.
Note: Students should compare equations correctly to get the exact values of requirements. And should solve the question step by step to avoid any calculation mistakes. They should know the value of $\tan {90^ \circ }$ is undefined so that they do not get confused while solving for $\alpha $.
Formula Used: $\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Complete step by step solution: Given, equations of curve $2{x^2} + 3{y^2} + 10x = 0$ and $3{x^2} + 5{y^2} + 16x = 0$
The equation of the curve through the point of intersection of the given curves
$2{x^2} + 3{y^2} + 10x + \lambda (3{x^2} + 5{y^2} + 16x) = 0$ ….(1)
We know If the equation of the line through the origin, then the equation must be homogeneous equation of second degree
equation must be of type ${x^2} + {y^2} = 0$
Therefore, the coefficient of $x$ is zero.
So, $10 + 16\lambda = 0$
$16\lambda = - 10$
$\lambda = \dfrac{{ - 5}}{8}$
Putting the value of $\lambda $ in equation (1)
$2{x^2} + 3{y^2} - \dfrac{5}{8}(3{x^2} + 5{y^2}) = 0$
$16{x^2} + 24{y^2} - 15{x^2} - 25{y^2} = 0$
${x^2} - {y^2} = 0$
The general equation is $a{x^2} + 2hxy + b{y^2} = 0$
On comparing, we will get
$a = 1$, $h = 0$ and $b = - 1$
We know the formula for angle between the two lines is
$\tan \alpha = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
$\tan \alpha = \left| {\dfrac{{2\sqrt {0 - (1)( - 1)} }}{{1 + ( - 1)}}} \right|$
After solving, we get
$\tan \alpha = \dfrac{2}{0}$
Taking ${\tan ^{ - 1}}$ on both side.
${\tan ^{ - 1}}\left( {\tan \alpha } \right) = {\tan ^{ - 1}}\left( {\dfrac{2}{0}} \right)$
$\alpha = {90^ \circ }$
Hence, the angle between the lines is ${90^ \circ }$
Therefore, option C is correct.
Note: Students should compare equations correctly to get the exact values of requirements. And should solve the question step by step to avoid any calculation mistakes. They should know the value of $\tan {90^ \circ }$ is undefined so that they do not get confused while solving for $\alpha $.
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