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  • The length of the perpendicular from the origin, on the normal to the curve,  at the point (2,2) is ; Opt

The length of the perpendicular from the origin, on the normal to the curve, x^{2}+2xy-3y^{2}=0 at the point (2,2) is ;
Option: 1 2
Option: 2 2\sqrt{2}
Option: 3 4\sqrt{2}  
Option: 4\sqrt{2}
 

Answers (1)

best_answer

 

 

Line parallel and perpendicular to a given line -

Line parallel and perpendicular to a given line

The equation of the line parallel to ax + by + c = 0 is given as ax + by + λ = 0, where λ is some constant. 

Equation of the given line is ax + by + c = 0 

Its slope is (-a/b)

So, any equation of line parallel to ax + by + c = 0 is 

The equation of the line perpendicular to ax + by + c = 0 is given as bx - ay + λ = 0, where λ is some constant. 

Equation of the given line is ax + by + c = 0 

Its slope is (-a/b)

Slope of perpendicular line will be (b/a)    

So, any equation of line perpendicular to ax + by + c = 0 is 

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Homogeneous Equations in Two Variables -

Homogeneous Equations in Two Variables

Homogeneous equations are those equations where each term has the same degree.

The equation ax2 + 2hxy + by2 = 0 is a homogeneous equation of second degree, it represents two straight lines through the origin.

\\\mathrm{The\;given\;equation\;is :}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;ax^2+2hxy+by^2=0\;\;\;\;\;\;\;\;\;\;\ldots(i)}\\\mathrm{Divide\;the \;equation\;by\;x^2}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;a+2h\left (\frac{y}{x} \right )+b\left (\frac{y^2}{x^2} \right )=0}\\\text{now put }\frac{y}{x} = m\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;a+2h\left (m \right )+b\left (m^2 \right )=0\;\;\;\;\;\ldots(ii)}\\\mathrm{If\;m_1\;and\;m_2\;are\;roots,\;then}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m_1+m_2=-\frac{2h}{b}}\\\mathrm{amd,\;\;\;\;\;\;\;\;m_1m_2=\frac{b}{a}}\\\mathrm{Thus,\;y=m_1x\;\;and\;\;y=m_2x\;are\;two\;straight\;lines\;given\;by\;Eq.\;(i)}
\\\text{Also, from the eq (iii)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m = \frac{-2h\pm\sqrt{\left ( h^2-ab \right )}}{2b}=\frac{y}{x}}\\\mathrm{\therefore \;\;\;\;\;\;\;\;\;\;\;\;by = \left \{ -h+\sqrt{(h^2-ab)} \right \}x}\\\mathrm{and \;\;\;\;\;\;\;\;\;by = \left \{ -h-\sqrt{(h^2-ab)} \right \}x}\\\text{are two lines represented by Eq. (i)}

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Distance of a Point From a Line -

Distance of a point from a line

Perpendicular length from a point (x1,y1) to the line L : Ax + By + C = 0 is 

-

\\ {x^{2}+2 x y-3 y^{2}=0} \\ {x^{2}+3 x y-x y-3 y^{2}=0} \\ {(x-y)(x+3 y)=0} \\ {x-y=0 \quad x+3 y=0}

(2, 2) satisfy x – y = 0

Normal 

x + y = \lambda= 4 Hence

the perpendicular distance from the origin  =\left|\frac{0+0-4}{\sqrt{2}}\right|=2 \sqrt{2}

Correct Option (2)

Posted by

vishal kumar

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