The product of the lengths of the perpendicul | vedclass

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

Question

(A) The given equation of the pair of lines is $x^2 + 3y^2 + 4xy - 4x - 10y + 3 = 0$. Comparing this with the general form $ax^2 + by^2 + 2hxy + 2gx +

Vedclass · Accepted Answer

$\frac{3}{\sqrt{20}}$

Vedclass · Answer

(A) The given equation of the pair of lines is $x^2 + 3y^2 + 4xy - 4x - 10y + 3 = 0$. Comparing this with the general form $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$,we get $a = 1, b = 3, h = 2, g = -2, f = -5, c = 3$. The product of the lengths of the perpendiculars from the origin $(0, 0)$ to the pair of lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by the formula $\frac{|c|}{\sqrt{(a-b)^2 + (2h)^2}}$. Substituting the values,we get $\frac{|3|}{\sqrt{(1-3)^2 + (2 	imes 2)^2}} = \frac{3}{\sqrt{(-2)^2 + 4^2}} = \frac{3}{\sqrt{4 + 16}} = \frac{3}{\sqrt{20}}$.

The product of the lengths of the perpendiculars from the origin to the pair of lines $x^2 + 3y^2 + 4xy - 4x - 10y + 3 = 0$ is

Similar Questions

If $2x^2 + 3xy - 2y^2 - 5x + 2fy - 3 = 0$ represents a pair of straight lines,then one of the possible values of $f$ is:

If the equation $x^2 + y^2 + 2gx + 2fy + 1 = 0$ represents a pair of straight lines,then:

If one of the lines in the pair of straight lines given by $4x^2+6xy+ky^2=0$ bisects the angle between the coordinate axes,then $k \in$

If $ax^2-xy-3y^2-5x+20y+c=0$ represents a pair of lines passing through the point $(2,3)$,then $a-c=$

$A$ homogeneous equation of second degree in $x$ and $y$ represents which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module