Tag: graphs to solve linear and non linear equations

Questions Related to graphs to solve linear and non linear equations

Let $C$ be a curve which is locus of the point of the intersection of lines $x=2+m$ and $my=4-m$. A circle $s\equiv (x-2)^{2}+(y+1)^{2}=25$ intersector the curve cut at four points $P,Q,R$ and $S$. If $O$ is centre of the curve $C$ the $OP^{2}+OQ^{2}+OR^{2}+OS^{2}$ is

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $50$

  2. $100$

  3. $25$

  4. $\dfrac{25}{2}$

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Correct Option: A

The point of intersection of the tangents drawn to the curve $x^2y=1 -y$ at the point where it is met by the curve xy=1-y is given by 

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. (0,-1)

  2. (1,1)

  3. (0,1)

  4. $(0,\infty )$

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Correct Option: A

If the lines joining the origin to the inter section of the line y = mx+2 and the curve ${ x }^{ 2 }+{ y }^{ 2 }=1$ are at right angles, then

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. ${ m }^{ 2 }=1$

  2. ${ m }^{ 2 }=3$

  3. ${ m }^{ 2 }=7$

  4. ${ 2m }^{ 2 }=1$

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Correct Option: A

If the line $y = \displaystyle \sqrt{3}x$ intersects the curve $\displaystyle x^{3}+y^{3}+3xy+5x^{2}+3y^{2}+4x+5y-1=0$ at the points $A, B, C,$ then the value of $OA.OB.OC$ is equal to: (here O is origin)

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $\displaystyle \frac{4}{13}\left ( 3\sqrt{3}+1 \right )$

  2. $\displaystyle \frac{4}{13}\left ( 3\sqrt{3}-1 \right )$

  3. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}-1 \right )$

  4. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}+1 \right )$

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Correct Option: B
Explanation:

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


The coordinates of these points can be written as ,

$A(x _1, \sqrt{3}x _1)$

$B(x _2, \sqrt{3}x _2)$

$C(x _3, \sqrt{3}x _3)$

If $O (0,0)$ is the origin then $OA = \sqrt { (x _1)^2 + (\sqrt{3x _1})^2 }$

$\Rightarrow OA = 2x _1$

Similarly  $OB = 2x _2$

and $OC = 2x _3$

Hence $OA .OB.OC = 8  \ (x _1.x _2.x _3)$

Now putiing value of $y = \sqrt3$ into equation of given curve, we get,

$ \Rightarrow x^3 + (\sqrt3x)^3 + 3.x.\sqrt3x + 5x^2 + 3 (\sqrt3x)^2 +4x - \sqrt3x -1 = 0$

$\Rightarrow ( 1 + 3\sqrt3)x^3 + (14 +3\sqrt3)x^2 + (4 -\sqrt3)x - 1=0$ ...$(1)$

The equation $(1)$ contains the abscissa of the intersection points of the given line and curve, which are $x _1$ , $x _2$ and $x _3$

From equation $(1)$ we can see that the product of roots is $x _1.x _2.x _3  = - \left ( \dfrac { - 1}{ 1 + 3\sqrt3} \right ) = \dfrac{1}{1 + 3\sqrt3} = \dfrac { 1- 3\sqrt3}{-26}$

Hence $OA.OB.OC = 8(x _1.x _2.x _3) = 8 \times \dfrac {1 - 3\sqrt3}{-26}$

$\Rightarrow OA.OB.OC = \dfrac{4}{13} (3\sqrt3 - 1)$

So correct option is $B$.

The least integral value of $a$ for which the graphs of the functions $y = 2ax + 1$ and $\displaystyle y=(a-6)x^{2}-2$ do not intersect is:

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. -6

  2. -5

  3. 3

  4. 2

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Correct Option: B
Explanation:

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.


Putting the value of $y$ from equation of line into equation of given parabola, we get,

$\Rightarrow (2ax + 1) = (a-6)x^2 - 2$

$\Rightarrow (a-6)x^2  - (2a)x -3 = 0$ ...$(1)$

Equation $(1)$ is a quadratic equation in $x$. 

For no intersection of both given functions, the equation $(1)$ must not have any real solutions.

A quadratic equation have no real roots if the value of it's discriminant is less than zero.

Hence $D = b^2 - 4ac < 0 $

$\Rightarrow D = ((-2a)^2) - 4 \times (a-6) \times (-3) < 0$

$\Rightarrow  D = 4a^2 +12a -72 < 0$

$\Rightarrow (a +6)(a -3) <0$

Hence Value of $a$ for the graphs of given functions do not intersect lies between $(-6 ,3)$

So the least integral value will be $(-5)$. Correct answer is $A$.

The point of intersection of the two ellipse $x^2+2y^2-6x-12y+23=0$ and $4x^2+2y^2-20x-12y+35=0$

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. lie on a circle centered at $\displaystyle \left( \frac { 8 }{ 3 } ,3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } $

  2. lie on a circle centered at $\displaystyle \left( -\frac { 8 }{ 3 } ,-3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } $

  3. lie on a circle centered at $\displaystyle \left( 8 ,9 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 3 }  } $

  4. are not concyclic

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Correct Option: A
Explanation:

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 ${S} _{2}=0.$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$   ...(1)
For it to be a circle, choose $\lambda$ such that the coefficients of ${x}^{2}$ and ${y}^{2}$ are equal: $\Rightarrow \lambda+4=2\lambda+2$
$\therefore \lambda=2$
This gives the equation of the circle as $6\left( { x }^{ 2 }+{ y }^{ 2 } \right) -32x-36y+81=0$  {(using (1))}
$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 16 }{ 3 } x-6y+\frac { 27 }{ 2 } =0$ 
Its centre is $\displaystyle C\left( \frac { 8 }{ 3 } ,3 \right) $ and radius is $\displaystyle r=\sqrt { \frac { 64 }{ 9 } +9-\frac { 27 }{ 2 }  } =\frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } .$

The line $x+y=1$ meets the lines represented by the equation $y^{3}-xy^{2}-14x^{2}y+24x^{3}=0$ at the points $A, B, C$. If $O$ is the origin, then $OA^{2}+OB^{2}+OC^{2}$ is equal to

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $\dfrac{22}9$

  2. $\dfrac{85}{72}$

  3. $\dfrac{181}{72}$

  4. $\dfrac{221}{72}$

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Correct Option: D
Explanation:

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


$24{ x }^{ 3 }+14{ x }^{ 2 }\left( x-1 \right) -x{ \left( x-1 \right)  }^{ 2 }-{ \left( x-1 \right)  }^{ 3 }=0\ \Rightarrow 36{ x }^{ 3 }-9{ x }^{ 2 }-4x+1=0\ \Rightarrow \left( 3x-1 \right) \left( 3x+1 \right) \left( 4x-1 \right) =0\ \Rightarrow x=\cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 4 } $

$\Rightarrow A\left( \cfrac { 1 }{ 3 } ,\cfrac { 2 }{ 3 }  \right) ,B\left( -\cfrac { 1 }{ 3 } ,\cfrac { 4 }{ 3 }  \right) $ and $C\left( \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 4 }  \right) $

Hence,

${ OA }^{ 2 }+{ OB }^{ 2 }+{ OC }^{ 2 }=\cfrac { 1 }{ 9 } +\cfrac { 4 }{ 9 } +\cfrac { 1 }{ 9 } +\cfrac { 16 }{ 9 } +\cfrac { 1 }{ 16 } +\cfrac { 9 }{ 16 } =\cfrac { 221 }{ 72 } $

The points of intersection of the two ellipses $x^{2}+2y^{2}-6x-12y+23=0$ and $4x^{2}+2y^{2}-20x-12y+35=0$.

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. lie on a circle centred at $\left(\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  2. lie on a circle centred at $\left(-\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  3. lie on a circle centred at $(8, 9)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  4. are not cyclic.

Show answer
Correct Option: A
Explanation:

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 $\displaystyle \sqrt{\left ( \frac{8}{3} \right )^{2}+\left ( 3 \right )^{2}-\frac{81}{6}}$
   $\displaystyle =\sqrt{\frac{128+162-243}{18}}=\frac{1}{3}\sqrt{\frac{47}{2}}$

If the points of intersection of curves $\displaystyle C _{1}=\lambda x^{2}+4y^{2}-2xy-9x+3: : and: : C _{2}=2x^{2}+3y^{2}-4xy+3x-1 $ subtends a right angle at origin then the value of $\displaystyle \lambda $ is

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $19$

  2. $9$

  3. $-19$

  4. $-9$

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Correct Option: C
Explanation:
Given 
$C _{1} : \lambda x^2+4y^2-2xy-9x+3=0$

$C _{2} :  2x^2+3y^2-4xy+3x-1=0\Rightarrow 2x^2+3y^2-4xy-1=-3x$-----(1)

Putting (1) in $C _{1}$

$\lambda x^2+4y^2-2xy+3(2x^2+3y^2-4xy-1)+3=0$

$\lambda x^2+4y^2-2xy+6x^2+9y^2-12xy-3+3=0$

$(\lambda+6 )x^2+13y^2-14xy=0$

Above equation has a condition of perpendicularity 

Hence $\lambda+6+13=0 \Rightarrow \lambda=-19$

If $x^{2}+y^{2}=a^{2}$ touches the line $y=3x+10$, then $a=$ 

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $\sqrt{10}$

  2. $10$

  3. $\sqrt{5}$

  4. $\dfrac{10}{2}$

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Correct Option: A