If the lines represented by the equation (p - | vedclass

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

Question

(D) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.For these lines to be mutually pe

Vedclass · Accepted Answer

$p$ and $q$ may have any value

Vedclass · Answer

(D) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.For these lines to be mutually perpendicular,the condition is $a + b = 0$.Comparing the given equation $(p - q)x^2 + 2(p + q)xy + (q - p)y^2 = 0$ with the general equation,we have $a = (p - q)$ and $b = (q - p)$.Substituting these into the condition $a + b = 0$,we get $(p - q) + (q - p) = 0$,which simplifies to $0 = 0$.This identity holds true for all values of $p$ and $q$. Therefore,$p$ and $q$ may have any value.

If the lines represented by the equation $(p - q)x^2 + 2(p + q)xy + (q - p)y^2 = 0$ are mutually perpendicular,then:

Similar Questions

The combined equation of the two diameters of a circle which divide the circle into $4$ sectors is $ax^2+2hxy+by^2=0$. If the area of the bigger sector is $5$ times the area of the smaller sector,then $\frac{|a+b|}{\sqrt{(a-b)^2+4h^2}} = $

If the acute angle between the lines given by $ax^2+2hxy+by^2=0$ is $\frac{\pi}{4}$,then $4h^2=$

The angle between the pair of straight lines $y^2 \sin^2 \theta - xy \sin^2 \theta + x^2(\cos^2 \theta - 1) = 0$ is

The angle between the lines given by $x^2 - y^2 = 0$ is ..... $^o$.

The angle between the pair of straight lines $3y^2 - 8xy - 3x^2 - 29x + 3y - 18 = 0$ is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module