The distance between the pair of parallel lin | vedclass

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

Question

(B) The given equation is $x^2+2xy+y^2-8ax-8ay-9a^2=0$. Comparing this with the general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=

Vedclass · Accepted Answer

$5 \sqrt{2} a$

Vedclass · Answer

(B) The given equation is $x^2+2xy+y^2-8ax-8ay-9a^2=0$. Comparing this with the general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=1, B=1, H=1, G=-4a, F=-4a, C=-9a^2$. The distance $d$ between two parallel lines represented by $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ is given by the formula $d = 2 \sqrt{\frac{G^2-AC}{A(A+B)}}$. Substituting the values: $d = 2 \sqrt{\frac{(-4a)^2 - (1)(-9a^2)}{1(1+1)}}$ $d = 2 \sqrt{\frac{16a^2 + 9a^2}{2}}$ $d = 2 \sqrt{\frac{25a^2}{2}} = 2 	imes \frac{5a}{\sqrt{2}} = 5\sqrt{2}a$. Thus,the distance is $5\sqrt{2}a$ units.

The distance between the pair of parallel lines represented by $x^2+2xy+y^2-8ax-8ay-9a^2=0$ is $...$ units.

Similar Questions

Find the equation of the pair of lines joining the origin to the points of intersection of the circle $x^2 + y^2 = 4$ and the line $2x + 3y - 4 = 0$.

The lines joining the points of intersection of the line $x + y = 1$ and the curve $x^2 + y^2 - 2y + \lambda = 0$ to the origin are perpendicular. Then the value of $\lambda$ is:

If the pair of lines joining the origin to the points of intersection of the line $x+y=1$ with the curve $x^2+y^2+2hxy+gx+fy+1=0$ are at right angles,then the point $(g, f)$ lies on the line

The lines joining the origin to the points of intersection of the curves $ax^2 + 2hxy + by^2 + 2gx = 0$ and $a'x^2 + 2h'xy + b'y^2 + 2g'x = 0$ will be mutually perpendicular,if

The angle between the lines joining the origin to the points of intersection of the straight line $y = 3x + 2$ with the curve $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module