If the equation ax^{2}+2hxy+by^{2}+2gx+2fy=0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the pair of lines is $ax^{2}+2hxy+by^{2}+2gx+2fy=0$. Since one of the lines is the bisector of the angle between the coordinate ax

Vedclass · Accepted Answer

$(a+b)^{2}=4(h^{2}+g^{2})$

Vedclass · Answer

(A) The equation of the pair of lines is $ax^{2}+2hxy+by^{2}+2gx+2fy=0$. Since one of the lines is the bisector of the angle between the coordinate axes,its equation is $y=x$ or $y=-x$,which can be written as $x-y=0$ or $x+y=0$. Let the other line be $lx+my+n=0$. Case $1$: If the line is $x-y=0$,then $(x-y)(lx+my+n) = lx^{2} + (m-l)xy - my^{2} + nx - ny = 0$. Comparing this with the given equation $ax^{2}+2hxy+by^{2}+2gx+2fy=0$,we get $a/l = 2h/(m-l) = b/(-m) = 2g/n = 2f/(-n)$. From $2g/n = 2f/(-n)$,we get $g = -f$. Also,$b/(-m) = 2f/(-n) \Rightarrow m = bn/(2f)$. Substituting these into the coefficients,we find the condition $(a+b)^{2} = 4(h^{2}+g^{2})$ or $4(h^{2}+f^{2})$. Given the options,the correct relation is $(a+b)^{2}=4(h^{2}+g^{2})$.

If the equation $ax^{2}+2hxy+by^{2}+2gx+2fy=0$ has one line as the bisector of the angle between the coordinate axes,then

Similar Questions

The difference of the slopes of the lines represented by the equation ${x^2}(\sec^2 \theta - \sin^2 \theta) - 2xy \tan \theta + y^2 \sin^2 \theta = 0$ is:

The equations of the lines represented by the equation ${x^2} - 5xy + 6{y^2} = 0$ are

If the slope of one of the lines given by $K x^2 + 6 x y + y^2 = 0$ is three times the other,then the value of $K$ is:

If the slopes of the lines represented by $ax^{2} + 2hxy + by^{2} = 0$ are in the ratio $1 : 3$,then $h^{2} : ab =$

The lines represented by the equation $x^2-y^2-x+3y-2=0$ are :

Vedclass Test Series

Exam Paper Generator

Online Exam Module