Tag: sets, relations and functions

$y^{2}+(m _{1}+m _{2})xy+m _{2}m _{1}x^2=0$

Comparing coefficients we get

$m _{1}+m _{2}=4$

$m _{1}m _{2}=1$

This implies

$m _{1}^{2}+1=4m _{1}$

$m _{1}^{2}-4m _{1}+1=0$

Therefore $m _{1}=2-\sqrt{3}$ and $m _{1}=2+\sqrt{3}$

$\tan A=2-\sqrt{3}$ and $\tan A=2+\sqrt{3}$

Hence $A=15^0$ and $A=75^0$

Corresponding values of $m _{2}=75^{\circ}$ and $15^{\circ}$

Therefore the angle between the lines is $75^0-15^0$

$=60^0$

The equation of the angle bisectors of the lines is $x=\pm y$

The line $x=-y$ is perpendicular to $x-y=4$

Hence the above is an isosceles triangle with the vertical angle being $60^0$

Hence the triangle is an equilateral triangle.

the $x$ and $y$ co-ordinates will be same 

$\Rightarrow h=-\dfrac{c}{2a}----------(1)$

Now,

$\ x+2y-10=0....(i)\ 2x+y+5=0....(ii)\times 2\ =>4x+2y+10=0....(iii)$

Subtracting (iii) and (i), we get,

$-3x-20=0\ =>x=\cfrac { -20 }{ 3 } \ \therefore y=\cfrac { 25 }{ 3 } $

Now the line must pass through $(\cfrac { -20 }{ 3 } ,\cfrac { 25 }{ 3 } )$ 

Therefore, $5x+4y=0$ passes through $(\cfrac { -20 }{ 3 } ,\cfrac { 25 }{ 3 } )$

Line $ = y - \sqrt 3 x + 3 = 0$

$\begin{array}{l} y=\sqrt { 3 } x-3 \\ m=\sqrt { 3 } =\tan  \theta  \\ \theta =\frac { \pi  }{ 3 } =60^{ \circ  }\to \left( i \right)  \end{array}$

Parametric form of line

$\begin{array}{l} =\frac { { x-{ x _{ 1 } } } }{ { \cos  \theta  } } =\frac { { y-{ y _{ 1 } } } }{ { \sin  \theta  } } =r  \\ \therefore x={ x _{ 1 } }+r \cos  \theta  \\ y={ y _{ 1 } }+r \sin  \theta  \end{array}$

As we know $P\left( {\sqrt 3 ,0} \right)$

$\therefore \left. \begin{array}{l} x=\sqrt { 3 } +r \cos  \theta  \\ y=0+r \sin  \theta  \end{array} \right\} lie\, \, on\, \, parabola\, \, { y^{ 2 } }=x+2$

$\begin{array}{l} \therefore { \left( { r\sin  \theta  } \right) ^{ 2 } }=\sqrt { 3 } +r\cos  \theta +2 \\ { r^{ 2 } }{ \sin ^{ 2 }  }\theta =\sqrt { 3 } +r\cos  \theta +2 \\ { r^{ 2 } }{ \sin ^{ 2 }  }\theta -r\cos  \theta -\sqrt { 3 } -2=0\, \, \, \begin{array} { *{ 20 }{ c } }{ { r _{ 1 } }=PA } \\ { { r _{ 2 } }=PB } \end{array} \\ { r _{ 1 } }{ r _{ 2 } }=\frac { { -\sqrt { 3 } -2 } }{ { { { \sin   }^{ 2 } }\theta  } } \to \left( { ii } \right)  \\ from\, \, \left( i \right) \, \, \theta =60^{ \circ  }\, \, \sin  60^{ \circ  }=\frac { { \sqrt { 3 }  } }{ 2 } \, \, { \sin ^{ 2 }  }60=\frac { 3 }{ 4 }  \\ from\, \, \left( { ii } \right) \, \, \frac { { -\sqrt { 3 } -2 } }{ { \frac { 3 }{ 4 }  } } =\frac { { -4\left( { \sqrt { 3 } +2 } \right)  } }{ 3 }  \\ \therefore \left| { PA.PB } \right| =\left| { { r _{ 1 } }{ r _{ 2 } } } \right| =\frac { { 4\left( { \sqrt { 3 } +2 } \right)  } }{ 3 }  \end{array}$

Hence ans is A.

$x+y=\left | a\right |$ and $ax-y=1$

Case $1:a>0$

$x+y=a-----(1)$

and $ax-y=1------(2)$

Adding $(1)+(2)$

$\Rightarrow x(1+a)=1+a$

$\Rightarrow x=1$

Hence, $y=a-1$

Since it is in first quadrant,$a-1\ge 0$

$\Rightarrow a\ge 1$

Case $2:a<0$

$x+y=-a$ and $ax-y=1$

Solving For $x,y$

$x=\dfrac{1-a}{1+a}>0$

$\Rightarrow \dfrac{a-1}{a+1}<0$

$\Rightarrow a\epsilon (-1,1)$

also ,$y=-a-\left(  \dfrac{1-a}{1+a}\right )$ which should be $>0$

$\Rightarrow -\dfrac{a^2+1}{a+1}>0$

$\Rightarrow a<-1$

Combining both two cases, we get :

$a\ge 1$

$a\epsilon[1,\infty)$

Given that the line $y-1=m(x-1)$ intersects the circle at two different points

If the perpendicular distance from the centre of the circle to the line is less than the radius of the circle then the line intersects at two different real points

$\Longrightarrow \dfrac { \left| m-1 \right|  }{ \sqrt { 1+{ m }^{ 2 } }  } <2\ $ 

Squaring on both sides gives,

${ m }^{ 2 }-2m+1<4+4{ m }^{ 2 }\ \Longrightarrow 3{ m }^{ 2 }+2m+3>0\ $,

Given quadratic equation has complex roots and the co-efficient of ${ x }^{ 2 }$ is positive

$\therefore$ The quadratic equation is always positive

Hence, infinite values of $m$ exist to intersect the line at $2$ diffferent real points.

$y=\left |x\right |+c$---(1)

$x^2+y^2-8\left| x\right |-9=0$

From equation (1) and (2)

$x^2+(\left | x\right |+c)^2-8\left |x\right |-9=0$

when $x>0$

$x^2+(x+c)^2-8x-9=0$

$x^2+x^2+c^2+2cx-8x-9=0$

$2x^2+x(2c-8)+c^2-9=0$

For no solution 

$D<0$

$(2c-8)^2-4\times 2 (c^2-9)<0$

$4c^2+64-32c-8c^2+72<0$

$-4c^2-32c+136<0$

$c^2+8c-34>0$

Hence for no solution c has all value excluding it's roots  

$x^2+y^2=5$

$\Rightarrow x^2+(x+1)^2=5$

$\Rightarrow x^2+x^2+1+2x=5$

$\Rightarrow 2x^2+1+2x-5=0$

$\Rightarrow 2x^2+2x-4=0$ or $x^2+x-2=0$

Factorising the above quadratic equation, we get:

$x^2+x-2=0$

$\Rightarrow x^2+2x-x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow (x+2)=0$ and $(x-1)=0$ 

$\rightarrow x=-2$ and $x=1$ 

Hence, the coordinates of the points intersection is $-2,1$.