The distance between the parallel lines 9x^2 | vedclass

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

Question

(D) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$. We can rewrite the quadratic part as $(3x - y)^2 + 6(3x - y) + 8 = 0$. Let $t = 3x -

Vedclass · Accepted Answer

$\frac{2}{\sqrt{10}}$

Vedclass · Answer

(D) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$. We can rewrite the quadratic part as $(3x - y)^2 + 6(3x - y) + 8 = 0$. Let $t = 3x - y$. Then the equation becomes $t^2 + 6t + 8 = 0$. Factoring the quadratic,we get $(t + 4)(t + 2) = 0$. Thus,the two parallel lines are $3x - y + 4 = 0$ and $3x - y + 2 = 0$. The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$. Here,$a = 3$,$b = -1$,$c_1 = 4$,and $c_2 = 2$. $d = \frac{|4 - 2|}{\sqrt{3^2 + (-1)^2}} = \frac{2}{\sqrt{9 + 1}} = \frac{2}{\sqrt{10}}$.

The distance between the parallel lines $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$ is

Similar Questions

The joint equation of the pair of lines passing through the origin and making an angle of $\frac{\pi}{4}$ with the line $3x + y - 6 = 0$ is

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

If in the general quadratic equation $f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,$\Delta = 0$ and $h^2 = ab$,then the equation represents:

The equation $x^2 - 7xy + 12y^2 = 0$ represents

If $L{x^2} - 10xy + 12{y^2} + 5x - 16y - 3 = 0$ represents a pair of straight lines,then the value of $L$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module