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Maths-

General

Easy

Question

The line among the following that touches the $y to the power of 2 end exponent equals 4 a x$ is

$x + m y + a m^{3} = 0$
$x - m y + a m^{2} = 0$
$x + m y - a m^{2} = 0$
$y + m x + a m^{2} = 0$

Hint:

The equation of tangent to parabola is given as $y equals m x space plus space a over m$ with slope = m.
Similarly find the equation of tangent to parabola with slope = $1 over m$ and simplify.

The correct answer is: $x minus m y plus a m to the power of 2 end exponent equals 0$

Condition 1
The line with slope 'm' is a tangent to parabola $y to the power of 2 end exponent equals 4 a x$
The equation of tangent to parabola is given as $y equals m x space plus space a over m$ (slope = m)
Condition 2
If slope = $1 over m$ , then equation of tangent
$rightwards double arrow y equals 1 over m x space plus space fraction numerator a over denominator begin display style bevelled 1 over m end style end fraction space equals space x over m space plus space m a rightwards double arrow y equals space fraction numerator x space plus space m squared a over denominator m end fraction rightwards double arrow y m space equals space x space plus m squared a space$

Thus, equation of lines touching parabola with slope $1 over m$ is $x minus m y plus a m to the power of 2 end exponent equals 0$ .

Related Questions to study

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In how many ways can six different rings be wear in four fingers?

Detailed Solution
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.

Now, we know that by the fundamental principle of counting there can be

4 \times 4 \times 4 \times 4 \times 4 \times 4

ways of wearing 6 rings.
So, we have

4 to the power of 6

4^{6} = 4096

ways to wear 6 different types of rings.

In how many ways can six different rings be wear in four fingers?

Detailed Solution
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.

Now, we know that by the fundamental principle of counting there can be

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So, we have

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parallel

A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

DETAILED SOLUTION:

There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.

Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
The number of ways of choosing 6 people out of 10 are

C presuperscript 10 subscript 6

,
Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are

C presuperscript 4 subscript 4 space end subscript equals space 1

And now all the 16 people are placed in their seats according to the constraints.

Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is

(8! \times 8!)

.
So, a possible number of arrangements will be

rightwards double arrow C presuperscript 10 subscript 6 space end subscript cross times space C presuperscript 4 subscript 4 space end subscript cross times space 8 factorial thin space cross times 8 factorial

C presuperscript n subscript r space end subscript equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction S o comma space C presuperscript 10 subscript 6 space end subscript equals space fraction numerator 10 factorial over denominator 4 factorial space cross times 6 factorial end fraction equals space fraction numerator 10.9.8.7.6 factorial over denominator 4.3.2.1.6 factorial end fraction equals space 210 A n d space C presuperscript 4 subscript 4 space end subscript equals space fraction numerator 4 factorial over denominator 4 factorial thin space cross times 0 factorial end fraction equals space 1

So total number of arrangements is

= 210 \times (8! \times 8!)

A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

DETAILED SOLUTION:

There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.

Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats).
The number of ways of choosing 6 people out of 10 are

C presuperscript 10 subscript 6

,
Now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are

C presuperscript 4 subscript 4 space end subscript equals space 1

And now all the 16 people are placed in their seats according to the constraints.

Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is

(8! \times 8!)

.
So, a possible number of arrangements will be

rightwards double arrow C presuperscript 10 subscript 6 space end subscript cross times space C presuperscript 4 subscript 4 space end subscript cross times space 8 factorial thin space cross times 8 factorial

C presuperscript n subscript r space end subscript equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction S o comma space C presuperscript 10 subscript 6 space end subscript equals space fraction numerator 10 factorial over denominator 4 factorial space cross times 6 factorial end fraction equals space fraction numerator 10.9.8.7.6 factorial over denominator 4.3.2.1.6 factorial end fraction equals space 210 A n d space C presuperscript 4 subscript 4 space end subscript equals space fraction numerator 4 factorial over denominator 4 factorial thin space cross times 0 factorial end fraction equals space 1

So total number of arrangements is

= 210 \times (8! \times 8!)

If ^(m+n)P₂ = 56 and ^m–nP₂ = 12 then (m, n) equals-

If ^(m+n)P₂ = 56 and ^m–nP₂ = 12 then (m, n) equals-

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point $P$ on its axis to infinity is

W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript

equals negative U subscript P end subscript equals negative m V subscript P end subscript

equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses

Potential at point

P

will be obtained by in integration as given below. Let

d M

be the mass of small rings as shown

d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r

equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction

d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction

equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r

equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses

therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis

A thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$ . The work required to take a unit mass from point $P$ on its axis to infinity is

physics-General

W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript

equals negative U subscript P end subscript equals negative m V subscript P end subscript

equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses

Potential at point

P

will be obtained by in integration as given below. Let

d M

be the mass of small rings as shown

d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r

equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction

d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction

equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r

equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses

therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis

parallel

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