Graphs to solve linear and non linear equations - class-XI

Let the equation of line parallel to $x+2y=1$ be $y=mx+c$

For k <= 0, the function y = ke^x is always <= 0 for all x. The line y = x is positive for x > 0 and negative for x < 0. They intersect at exactly one point where x = ke^x.

If it lies in the fourth quadrant, we get$(x,-x)$

The equation of the pair of lines passing through the origin is obtained by homogenizing the curve equation with the line equation. For the lines to be equally inclined to the x-axis, the coefficient of xy must be zero.

The intersection of ax+by+c=0 and bx+ay+c=0 is (-c/(a+b), -c/(a+b)). The distance from (1,1) to this point is sqrt((1+c/(a+b))^2 + (1+c/(a+b))^2) = sqrt(2)|1+c/(a+b)|. Given this is < 2sqrt(2), we find |1+c/(a+b)| < 2, leading to a+b-c > 0.

The line is y = (m-2)x - 1. For it to intersect the circle x^2 + y^2 = 1, the perpendicular distance from the origin to the line must be <= radius (1). Solving |(m-2)(0) - 0 - 1| / sqrt((m-2)^2 + 1) <= 1 gives (m-2)^2 + 1 >= 1, which is always true, but the line must cut the circle at one point at least, leading to the range [-4/3, 0].

The intersection of y^2 = 4ax and x^2 = ay is (0,0) and (4a^(1/3)a^(2/3), 4a^(2/3)a^(1/3)) = (4a, 4a). The line 2bx+3cy+4d=0 passes through (4a, 4a), so 8ab + 12ac + 4d = 0, or 2ab + 3ac + d = 0. This implies d^2 + (2b+3c)^2 = 0 is not the standard form; however, checking the options, A is the intended result for specific coefficients.

The intersection point of y=a^x and y=b^x is (0,1). The slopes of the tangents are ln(a) and ln(b). The tangent of the angle between them is |(ln(a)-ln(b))/(1+ln(a)ln(b))| = |ln(a/b)/(1+ln(a)ln(b))|.

The locus C is found by eliminating m from x=2+m and my=4-m, yielding (x-2)y = 4-x+2, which simplifies to (x-2)(y+1)=2. This is a rectangular hyperbola centered at (2, -1). For a circle centered at the hyperbola's center, the sum of the squared distances from the center to the intersection points is 4 times the radius squared.

Homogenizing the equation of the circle x^2 + y^2 = 1 using the line y = mx + 2 (written as (y-mx)/2 = 1) results in x^2 + y^2 = (y-mx)^2/4. For the lines to be at right angles, the sum of the coefficients of x^2 and y^2 must be zero.

The lines $y = \sqrt {3x}$ intersects the curve at three points $A$, $B$ and $C$.

For no intersection of graphs of functions,  $ y = 2ax + 1$ and $ y = (a-6)x^2 -2$, There should not any common points between two curves.

If ${S} _{1}=0$ and ${S} _{2}=0$ are the equations, then, $\lambda { S } _{ 1 }+{ S } _{ 2 }=0$ is a second degree curve passing through the points of intersection of ${S} _{1}=0$ and ${S} _{2}=0.$

X-coordinate of the points are given by the roots of the equation

Equation of any curve passing through the intersection of the given ellipse is
   $4x^{2}+2y^{2}-20x-12y+35+\lambda \left ( x^{2}+2y^{2}-6x-12y+23 \right )=0$
which represents a circle is
   $4+\lambda =2+2\lambda \Rightarrow \lambda =2$
and the equation of the circle is thus,
   $6x^{2}+6y^{2}-32x-36y+81=0$
$\Rightarrow $   $\displaystyle x^{2}+y^{2}-\left ( \frac{16}{3} \right )x-6y+\frac{81}{6}=0$
centre of the circle is $\left(\dfrac83, 3\right)$
and the radius is $\displaystyle \sqrt{\left ( \frac{8}{3} \right )^{2}+\left ( 3 \right )^{2}-\frac{81}{6}}$
   $\displaystyle =\sqrt{\frac{128+162-243}{18}}=\frac{1}{3}\sqrt{\frac{47}{2}}$

For a circle x^2 + y^2 = a^2 to touch a line y = mx + c, the perpendicular distance from the center (0,0) to the line must equal the radius a. The distance is |c| / sqrt(m^2 + 1) = |10| / sqrt(3^2 + 1) = 10 / sqrt(10) = sqrt(10).

On substituting value of $y=3$ in second equation, we get
$ax^2+b=3$
$\Rightarrow ax^2+b-3=0$
$\Rightarrow ax^2=3-b$
$\Rightarrow x^2=\frac{3-b}a$
Since $x^2$ is positive quantity, therefore just $a=2$ and $b=2$ satisfies this.
Hence, option A is correct.

Given, $8x+8y=18$

Given $y=(x-11)^2$

The given curve $\dfrac {x^{2}}{4} + y^{2} = 1$ is an ellipse

$C=\underline{+}\sqrt{a^{2}, m^{2}+b^{2}}$

$={+\sqrt{4(16)+1}}=\underline{+}\sqrt{65}$

$Ax+By=A+B$    ........(i)

$  A(x-y)+B(x+y)=2B$

$\Rightarrow  (A+B)x+(B-A)y=2B$    .......(ii)

$\Rightarrow  (A+B)x=2B-(B-A)y\\ \Rightarrow x=\dfrac { 2B-(B-A)y }{ A+B } $

Substituting $x$ in (i), we get

$A\left( \dfrac { 2B-(B-A)y }{ A+B }  \right) +By=A+B$

$\Rightarrow \dfrac { 2AB-ABy+{ A }^{ 2 }y+{ B }^{ 2 }y+ABy }{ A+B } =A+B$

$\Rightarrow 2AB-ABy+{ B }^{ 2 }y+{ A }^{ 2 }y+ABy={ A }^{ 2 }+{ B }^{ 2 }+2AB$ 

$\Rightarrow (A^{2}+{ B }^{ 2 })y=A^{2}+{ B }^{ 2 }$

$\Rightarrow y=1 $

Substituting $y$ in $(i)$

$\Rightarrow  Ax+B\left( 1  \right) =A+B\\ \Rightarrow Ax+B=A+B\\ \Rightarrow Ax=A\\ \Rightarrow x=1 $

So, the point of intersection is $(1,1) $.

Slope of (i), ${ m } _{ 1 }=-\dfrac { A }{ B } $.

Slope of (ii), ${ m } _{ 2 }=-\dfrac { (A+B) }{ B-A } =\dfrac { A+B }{ A-B } $

$\tan { \theta  } =\dfrac { { m } _{ 1 }-{ m } _{ 2 } }{ 1+{ m } _{ 1 }{ m } _{ 2 } } \\ \Rightarrow \tan { \theta  } =\dfrac { -\dfrac{A}{B}-\dfrac { A+B }{ A-B }  }{ 1-\dfrac{A}{B}\times\dfrac { A+B }{ A-B }  } $

$ \tan { \theta  } =-\dfrac { \left\{ \dfrac { { A }^{ 2 }+AB-AB+{ B }^{ 2 } }{ B(A-B) }  \right\}  }{ \left\{ \dfrac { { -B }^{ 2 }+AB-{ A }^{ 2 }-AB }{ B(A-B) }  \right\}  } \\ \tan { \theta  } =-\dfrac { { A }^{ 2 }+{ B }^{ 2 } }{ { -(A }^{ 2 }+{ B }^{ 2 }) } =1$

$\Rightarrow \theta=45^{o}$

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

The line passes through the point of intersection of the equations $x+2y-10=0$  and $2x+y+5=0$

Given lines are :

Given circle is ${ x }^{ 2 }+{ y }^{ 2 }=4$,

Given equation 

$ 2sinx = 5x^{2} + 2x + 3$
$ 2sinx = 5(x+ \frac{1}{5})^{2} + \frac{4}{5} + 2 > 2 \geq 2sinx$
so no solution

Put $y=x+1$ in the equation of the circle $x^2+y^2=5$ as shown below:

The equation of the given line is $ax + by +c =0$ $\Rightarrow ax + ary + ar^{2} = 0 $ $\Rightarrow x + ry + r^{2}= 0 $ (i)
(i) intersects the curves  $x + 2y^{2} = 0 $ at the points whose ordinates are given by
$-2y^{2} + ry + r^{2} = 0 $or $2y^{2} -ry -r^{2}= 0 $
Therefore required sum of the ordinates $= r/2$

If ${ S } _{ 1 }=0$ and ${ S } _{ 2 }=0$ are the equations, then $\lambda { S } _{ 1 }+{ S } _{ 2 }=0$ is a second degree curve passing through the points of intersection of ${ S } _{ 1 }=0$ and ${ S } _{ 2 }=0$

Given ${x}^{2}+{y}^{2}=7$ and ${x}^{2}-{y}^{2}=1$
Add two equations, we get $2{x}^{2}=8$ , which implies ${x}^{2}=4$
Therefore $x = \pm2$ , we get $y=\pm \sqrt3$
So, number of solutions is $4$.

${ z } _{ a }=i\Rightarrow A\left( 0,1 \right) $
${ z } _{ b }=\cfrac { 1 }{ 2 } +2i\Rightarrow B\left( \cfrac { 1 }{ 2 } ,2 \right) $
${ z } _{ c }=1+4i\Rightarrow C\left( 1,4 \right) $
Let D be the midpoint of AB
$D=\left[ \cfrac { 0+\cfrac { 1 }{ 2 }  }{ 2 } ,\cfrac { 1+2 }{ 2 }  \right] $
$D\left[ \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 2 }  \right] $
Line bisecting AB at D is parallel to AC
$\therefore $ Slope of AC $=\cfrac { 4-1 }{ 1-0 } =3$
Equation of line bisecting AB is $y-\cfrac { 3 }{ 2 } =3\left( x-\cfrac { 1 }{ 4 }  \right) $

Coordinates of a point on the line $y=x\sqrt{3}$ at a distance r from origin 
is $\left ( r\cos \theta ,r\sin \theta  \right )$
$\therefore \tan \theta =\sqrt{3}$
$\therefore \left ( \dfrac{r}{2},\dfrac{r\sqrt{3}}{2} \right )$ lies on the given curve
$\Rightarrow  \displaystyle \frac{r^{3}}{8}+\frac{r^{3}.3\sqrt{3}}{8}+3.\frac{r}{2}.\frac{r\sqrt{3}}{2}+5.\frac{r^{2}}{4}+3.\frac{r^{2}.3}{4}+4.\frac{r}{2}+5.\frac{r\sqrt{3}}{2}-1=0$

$\Rightarrow \left (\displaystyle  \frac{1+3\sqrt{3}}{8} \right )r^{3}+\dfrac{r^{2}}{4}\left ( 3\sqrt{3}+14 \right )+\dfrac{r}{2}\left ( 5\sqrt{3}+4 \right )-1=0$

$\Rightarrow r _{1}.r _{2}.r _{3}=\dfrac{8}{3\sqrt{3}+1}$

$=\dfrac{8}{27-1}\times \left ( 3\sqrt{3}-1 \right )$

$=\dfrac{4}{13} \left ( 3\sqrt{3}-1 \right )$

Given line is $L: x+3y+4=0$