Tag: maths

Questions Related to maths

The difference between two angles is $19$$\displaystyle ^{o}$ and their sum is $\displaystyle \frac{890}{9}^o$. Find the greater angle.

maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

  1. $63^o$

  2. $35^o$

  3. $27^o$

  4. $59^o$

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Correct Option: D
Explanation:
Let the two angles be $a$ and $b$

Then, 
$a  - b = 19$

and, $a + b = \dfrac{890}{9}$

Adding the two equations,

$2a = \dfrac{890 + 19\times 9}{9}$

$2a = \dfrac{1061}{9}$

Thus, $a = \dfrac{1061}{18}$

$a \approx 59^{\circ}$

If two angles of a triangle are acute angles, the third angle:

maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

  1. is less than the sum of the two angles

  2. is an acute angle

  3. is the largest angle of the triangle

  4. may be an obtuse angle

Show answer
Correct Option: D
Explanation:
For a triangle $ABC$, sum of angles is $180^0$
Two angles are given as acute, $\angle A, \angle B < 90^0$
$\angle A = 90^0 - x$
$\angle B = 90^0 - y$
where, $x,y < 90^0$
$\therefore \angle A + \angle B+ \angle C = 180^0$
$\Rightarrow \angle C = x+y $
$x+y$ can be $>90^0$ or $<90^0$
So, it may be obtuse or acute.
Hence, option D.

An angle which measures $\displaystyle 0^{o}$ is called:

maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

  1. obtuse angle

  2. straight angle

  3. zero angle

  4. right angle

Show answer
Correct Option: C
Explanation:

An angle which measure $0^0$ is called zero angle.

An angle which measure more than $0^o$ and less than $90^o$ is an acute angle.
An angle which measures $90^o$ is called a right angle.
An angle which measure more than $90^o$ and less than $180^o$ is an obtuse angle.

Hence, the answer is zero angle.

Find the supplement  of the following angle.
$40^{\circ}$

maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

  1. $140$

  2. $40$

  3. $10$

  4. None of these

Show answer
Correct Option: A
Explanation:

We know that the supplement angle

$=180^0-\theta$

So,
The supplement angle of $40^0$ will be
$=180^0-40^0=140^0$ 

Hence, this is the answer.

The length of the chord of the parabola $y^2 = 4x$ which passes through the vertex and makes $30^o$ angle with x-axis is

intersection of a line and a parabola conic section maths

  1. $\dfrac{\sqrt{3}}{2}$

  2. $\dfrac{3}{2}$

  3. $8\sqrt{3}$

  4. $\sqrt{3}$

Show answer
Correct Option: C

If a$\ne $b then the length of common chord of the circles ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}$ and ${\left( {x - {b^{}}} \right)^2} + {\left( {y - a} \right)^2} = c^2$ is 

intersection of a line and a parabola conic section maths

  1. $\sqrt {{c^2} - {{\left( {a - b} \right)}^2}} $

  2. $\sqrt {4{c^2} - 2{{\left( {a - b} \right)}^2}} $

  3. $\sqrt {3{c^2} - {{\left( {a - b} \right)}^2}} $

  4. $\sqrt {2{c^2} - {{\left( {a - b} \right)}^2}} $

Show answer
Correct Option: B
Explanation:

We have,

${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{c}^{2}}$ 
${{S} _{1}}\equiv {{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{b}^{2}}-2by-{{c}^{2}}=0$               …….. (1)

 

${{\left( x-b \right)}^{2}}+{{\left( y-a \right)}^{2}}={{c}^{2}}$

${{S} _{2}}\equiv {{x}^{2}}+{{b}^{2}}-2bx+{{y}^{2}}+{{a}^{2}}-2ay-{{c}^{2}}=0$                        ……… (2)

 

Since, $a\ne b$

 

Centre of the circle ${{S} _{1}}=\left( a,b \right)$ and radius ${{r} _{1}}=c$.

 

We know that the equation of common chord is ${{S} _{1}}-{{S} _{2}}=0$

 

So, the equation is

$\left( b-a \right)x+\left( a-b \right)y=0$             …….. (3)

 

We know that the length of common chord is

$=2\sqrt{{{r} _{1}}^{2}-{{d} _{1}}^{2}}$                      ………. (4)

Where ${{r} _{1}}=$ radius and ${{d} _{1}}$ is the length of perpendicular drawn from the centre to the chord.

 

So,

${{d} _{1}}=\left| \dfrac{\left( b-a \right)a+\left( a-b \right)b}{\sqrt{{{\left( b-a \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right|$

$ {{d} _{1}}=\left| \dfrac{ab-{{a}^{2}}+ab-{{b}^{2}}}{\sqrt{{{\left( a-b \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{2ab-{{a}^{2}}-{{b}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-{{\left( a-b \right)}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-\left( a-b \right)}{\sqrt{2}} \right| $

$ {{d} _{1}}=\dfrac{\left( a-b \right)}{\sqrt{2}} $

 

From equation (4),

The length of common chord $ =2\sqrt{{{c}^{2}}-{{\left( \dfrac{a-b}{\sqrt{2}} \right)}^{2}}} $

$ =2\sqrt{{{c}^{2}}-{{\dfrac{\left( a-b \right)}{2}}^{2}}} $

$ =2\sqrt{{{\dfrac{2{{c}^{2}}-\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{{{\dfrac{8{{c}^{2}}-4\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{4{{c}^{2}}-2{{\left( a-b \right)}^{2}}} $

 

Hence, this is the answer.

Length of chord of parabola ${y}^{2}=4ax$ whose equation is $y-\sqrt {2}x+4\sqrt {2}a=0$

intersection of a line and a parabola conic section maths

  1. $2\sqrt {11}a$

  2. $4\sqrt {2}a$

  3. $8\sqrt {2}a$

  4. $6\sqrt {3}a$

Show answer
Correct Option: A

If a$\ne $b then the length of common chord of the circles ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}$ and ${\left( {x - {b^{}}} \right)^2} + {\left( {y - a} \right)^2} = c^2$ is 

intersection of a line and a parabola conic section maths

  1. $\sqrt {{c^2} - {{\left( {a - b} \right)}^2}} $

  2. $\sqrt {3{c^2} - {{\left( {a - b} \right)}^2}} $
  3. $\sqrt {4{c^2} - 2{{\left( {a - b} \right)}^2}} $

  4. $\sqrt {2{c^2} - {{\left( {a - b} \right)}^2}} $

Show answer
Correct Option: C
Explanation:

We have,

${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{c}^{2}}$ 
${{S} _{1}}\equiv {{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{b}^{2}}-2by-{{c}^{2}}=0$               …….. (1)

 

${{\left( x-b \right)}^{2}}+{{\left( y-a \right)}^{2}}={{c}^{2}}$

${{S} _{2}}\equiv {{x}^{2}}+{{b}^{2}}-2bx+{{y}^{2}}+{{a}^{2}}-2ay-{{c}^{2}}=0$                        ……… (2)

 

Since, $a\ne b$

 

Centre of the circle ${{S} _{1}}=\left( a,b \right)$ and radius ${{r} _{1}}=c$.

 

We know that the equation of common chord is ${{S} _{1}}-{{S} _{2}}=0$

 

So, the equation is

$\left( b-a \right)x+\left( a-b \right)y=0$             …….. (3)

 

We know that the length of common chord is

$=2\sqrt{{{r} _{1}}^{2}-{{d} _{1}}^{2}}$                      ………. (4)

Where ${{r} _{1}}=$ radius and ${{d} _{1}}$ is the length of perpendicular drawn from the centre to the chord.

 

So,

${{d} _{1}}=\left| \dfrac{\left( b-a \right)a+\left( a-b \right)b}{\sqrt{{{\left( b-a \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right|$

$ {{d} _{1}}=\left| \dfrac{ab-{{a}^{2}}+ab-{{b}^{2}}}{\sqrt{{{\left( a-b \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{2ab-{{a}^{2}}-{{b}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-{{\left( a-b \right)}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-\left( a-b \right)}{\sqrt{2}} \right| $

$ {{d} _{1}}=\dfrac{\left( a-b \right)}{\sqrt{2}} $

 

From equation (4),

The length of common chord $ =2\sqrt{{{c}^{2}}-{{\left( \dfrac{a-b}{\sqrt{2}} \right)}^{2}}} $

$ =2\sqrt{{{c}^{2}}-{{\dfrac{\left( a-b \right)}{2}}^{2}}} $

$ =2\sqrt{{{\dfrac{2{{c}^{2}}-\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{{{\dfrac{8{{c}^{2}}-4\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{4{{c}^{2}}-2{{\left( a-b \right)}^{2}}} $

 

Hence, this is the answer.

The length of the chord $y = x - 2$ intercepted by the parabola ${ y }^{ 2 }=4(x-1)$ is 

intersection of a line and a parabola conic section maths

  1. $4$

  2. $\dfrac { 16 }{ 3 } $

  3. $\dfrac { 3 }{ 16 } $

  4. $\dfrac { 1 }{ 4 } $

Show answer
Correct Option: A

The length of normal chord to the parabola $y^{2} = 4x$ which subtends a right angle at the vertex is

intersection of a line and a parabola conic section maths

  1. $6\sqrt {3}$

  2. $6\sqrt {2}$

  3. $7\sqrt {2}$

  4. $7\sqrt {3}$

Show answer
Correct Option: A
Explanation:

A chord is a line segment that passes through any two points on the parabola. A normal chord is a chord that is perpendicular to a tangent of the parabola at the point of intersection of the chord with the parabola. 


Let $y=mx+c$   is the tangent of parabola.


As given parabola is $y^2=4x$

So any point on the parabola is $(t^2,2t)$

As $y=mx+\dfrac{a}{m}$ , is the tangent to the parabola $y^2=4ax$

then the tangent equation to this given parabola is $y=mx+\dfrac{1}{m}$

Let the tangent passes through point $P(t _1^{2},2t _1)$

On substituting P in parabola equation we get $m=\dfrac{1}{t _1}$

So the slope of the normal is $-t _1$

Let the chord joins $P(t _1^2,2t _1)$ and $Q(t _2^2,2t _2)$ 

On solving we will get slope of line PQ as $\dfrac{2}{t _1+t _2}$

So , $\dfrac{2}{{t _1}+{t _2}}= \dfrac{1}{t _1}$

$\Rightarrow t _1^{2}+{t _1}{t _2}=-2$

As from properties of a normal chord which subtends a right angle at the vertex, ${t _1}{t _2}=-4$

On solving above two equations we get $ t _1=\sqrt{2} , t _2=-2\sqrt{2} $

Hence the points are $P(2,2\sqrt{2})$ and $Q(8,-4\sqrt{2}) $

By applying distance formula we get the distance between P and Q as

$\Rightarrow PQ=\sqrt{(8-2)^2+(-4\sqrt{2}-2\sqrt{2})^2}$

$\Rightarrow PQ=\sqrt{108}=6\sqrt{3} units $