The square of the distance from the origin to | 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 $ax^2+2hxy-ay^2+2gx+2fy+c=0$. Let the point of intersection be $(x_0, y_0)$. The coordinates of the poi

Vedclass · Accepted Answer

$\frac{f^2+g^2}{a^2+h^2}$

Vedclass · Answer

(A) The given equation of the pair of lines is $ax^2+2hxy-ay^2+2gx+2fy+c=0$. Let the point of intersection be $(x_0, y_0)$. The coordinates of the point of intersection of the pair of lines $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ are given by the formulas: $x_0 = \frac{HF-BG}{AB-H^2}$ and $y_0 = \frac{GH-AF}{AB-H^2}$. Here,$A=a$,$H=h$,$B=-a$,$G=g$,$F=f$,$C=c$. Substituting these values: $x_0 = \frac{hf-g(-a)}{a(-a)-h^2} = \frac{hf+ag}{-(a^2+h^2)}$ $y_0 = \frac{gh-af}{-(a^2+h^2)} = \frac{af-gh}{a^2+h^2}$. The square of the distance from the origin $(0,0)$ is $x_0^2 + y_0^2$. $x_0^2 + y_0^2 = \frac{(hf+ag)^2 + (af-gh)^2}{(a^2+h^2)^2} = \frac{h^2f^2+a^2g^2+2afgh + a^2f^2+g^2h^2-2afgh}{(a^2+h^2)^2}$ $= \frac{f^2(h^2+a^2) + g^2(a^2+h^2)}{(a^2+h^2)^2} = \frac{(f^2+g^2)(a^2+h^2)}{(a^2+h^2)^2} = \frac{f^2+g^2}{a^2+h^2}$.

The square of the distance from the origin to the point of intersection of the pair of lines $ax^2+2hxy-ay^2+2gx+2fy+c=0$ is

Similar Questions

The point of intersection of the lines represented by the equation $2x^2 + 7xy + 3y^2 + 8x + 14y + 8 = 0$ is

If the pairs of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 q x y-y^2=0$ are such that each pair bisects the angle between the other pair,then

The joint equation of the pair of lines which bisects the angles between the lines $x^2+3xy+2y^2=0$ is

If the bisectors of the angles represented by $ax^2 + 2hxy + by^2 = 0$ and $a'x^2 + 2h'xy + b'y^2 = 0$ are the same,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module