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Question

The points of contact of the vertical tangents to $x equals 2 minus 3 blank s i n blank theta comma$ $y equals 3 plus 2 blank c o s blank theta$ are

(2,5), (2,1)
(-1,3), (5,3)
(2,5)(5,3)
(-1,3)(2,1)

hint Hint:

A normal in geometry is a piece of geometry that is perpendicular to another piece of geometry, such as a line, ray, or vector. For instance, the (infinite) line perpendicular to the tangent line to the curve at the point is the normal line to a plane curve at the point. Here we have to find the points of contact of the vertical tangents to $x equals 2 minus 3 blank s i n blank theta comma$ $y equals 3 plus 2 blank c o s blank theta$ .

The correct answer is: (-1,3), (5,3)

At a specific point on a curve, tangents and normals behave just like any other straight lines. The normal runs perpendicular to the curve, and a tangent is parallel to the curve at the point. Like any other straight line, the equation of tangent and normal can be evaluated.
Here we have given equations as $x equals 2 minus 3 blank s i n blank theta comma$ $y equals 3 plus 2 blank c o s blank theta$ .

$N o w space f o r space t h e space v e r t i c a l space tan g e n t s comma space w e space h a v e colon fraction numerator d x over denominator d theta end fraction equals 0 comma s o space w i t h space t h i s space w e space g e t colon minus 3 cos theta equals 0 theta equals straight pi over 2 o r space fraction numerator 3 straight pi over denominator 2 end fraction N o w space w e space h a v e space t h e space c o r r e s p o n d i n g space v a l u e s space t o space theta comma space t h a t space a r e colon x equals 2 minus 3 sin straight pi over 2 equals negative 1 semicolon y equals 3 plus 2 cos straight pi over 2 equals 3 semicolon x equals 2 minus 3 sin fraction numerator 3 straight pi over denominator 2 end fraction equals 2 plus 3 equals 50 semicolon y equals 3 plus 2 cos fraction numerator 3 straight pi over denominator 2 end fraction equals 3 semicolon S o space a s space p e r space t h i s space t h e space p o i n t s space a r e space left parenthesis negative 1 comma 3 right parenthesis space a n d space left parenthesis 5 comma 3 right parenthesis.$

So here we used the concept of tangents and normals, we found the slope and then went to proceed for for the final answer. Using the vertical tangents we found equation and then the points.
$S o space a s space p e r space t h i s space t h e space p o i n t s space a r e space left parenthesis negative 1 comma 3 right parenthesis space a n d space left parenthesis 5 comma 3 right parenthesis.$ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mi>o</mi><mo> </mo><mi>a</mi><mi>s</mi><mo> </mo><mi>p</mi><mi>e</mi><mi>r</mi><mo> </mo><mi>t</mi><mi>h</mi><mi>i</mi><mi>s</mi><mo> </mo><mi>t</mi><mi>h</mi><mi>e</mi><mo> </mo><mi>p</mi><mi>o</mi><mi>i</mi><mi>n</mi><mi>t</mi><mi>s</mi><mo> </mo><mi>a</mi><mi>r</mi><mi>e</mi><mo> </mo><mo>(</mo><mo>-</mo><mn>1</mn><mo>,</mo><mn>3</mn><mo>)</mo><mo> </mo><mi>a</mi><mi>n</mi><mi>d</mi><mo> </mo><mo>(</mo><mn>5</mn><mo>,</mo><mn>3</mn><mo>)</mo><mo>.</mo></math>

The curve $y equals a x to the power of 3 end exponent plus b x to the power of 2 end exponent plus c x plus 5$ touches the $x$ ‐axis at P(-2,0) and cuts the y ‐axis at a point $Q left parenthesis O comma 5 right parenthesis$ where its gradient is 3 Then a+2b+c=-

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If $open vertical bar table row alpha beta row gamma cell negative alpha end cell end table close vertical bar$ is to be square root of a determinant of the two rowed unit matrix, then $alpha comma beta comma$ $gamma$ should satisfy the relation

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Matrix Ais given by $A equals open curly brackets table attributes columnalign left end attributes row 3 11 row 2 8 end table close curly brackets$ then the determinant of $A to the power of 2011 end exponent minus 5 A to the power of 2010 end exponent$ is

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If $A open parentheses table row 1 3 4 row 3 cell negative 1 end cell 5 row cell negative 2 end cell 4 cell negative 3 end cell end table close parentheses equals open parentheses table row 3 cell negative 1 end cell 5 row 1 3 4 row cell plus 4 end cell cell negative 8 end cell 6 end table close parentheses$ then $A equals negative times$

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If $open vertical bar table row cell a blank b blank 1 end cell row cell b blank c blank 1 end cell row cell c blank a blank 1 end cell end table close vertical bar equals 2010 text andif end text open vertical bar table row cell c minus a end cell cell c minus b end cell cell a b end cell row cell a minus b end cell cell a minus c end cell cell b c end cell row cell b minus c end cell cell b minus a end cell cell c a end cell end table close vertical bar minus open vertical bar table row cell c minus a end cell cell c minus b end cell cell c to the power of 2 end exponent end cell row cell a minus b end cell cell a minus c end cell cell a to the power of 2 end exponent end cell row cell b minus c end cell cell b minus a end cell cell b to the power of 2 end exponent end cell end table close vertical bar equals p$ then the number of positive divisors of the number p is

maths-General

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The number of positive integral solutions of the equation $open vertical bar table row cell y to the power of 3 end exponent plus 1 end cell cell y to the power of 2 end exponent z end cell cell y to the power of 2 end exponent x end cell row cell y z to the power of 2 end exponent end cell cell z to the power of 3 end exponent plus 1 end cell cell z to the power of 2 end exponent x end cell row cell y x to the power of 2 end exponent end cell cell X to the power of 2 end exponent Z end cell cell x to the power of 3 end exponent plus 1 end cell end table close vertical bar equals 11$ is

maths-General

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If $A comma$ $B comma$ $C comma$ $D$ be real matrices (not necessasrily square) such that $A to the power of T end exponent equals B C D comma$ $B to the power of T end exponent equals C D A comma$ $C to the power of T end exponent equals D A B comma$ $D to the power of T end exponent equals A B C$ forthematrix $S equals A B C D$ consider the statements $I colon S to the power of 3 end exponent equals S$ $capital pi colon S to the power of 2 end exponent equals S to the power of 4 end exponent$

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If $l o g subscript 4 end subscript left parenthesis x plus 2 y right parenthesis plus l o g subscript 4 end subscript left parenthesis x minus 2 y right parenthesis equals 1$ , then
Statement $negative 1 colon left parenthesis vertical line x vertical line minus vertical line y vertical line right parenthesis subscript blank m i n blank end subscript equals square root of 3$ because
Statement $negative 2$ :A.M $greater or equal than G. M$ for $R to the power of plus end exponent.$

maths-General

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maths-

If $alpha element of left parenthesis negative fraction numerator pi over denominator 2 end fraction comma 0 right parenthesis$ , then the value of $t a n to the power of negative 1 end exponent left parenthesis blank c o t blank alpha right parenthesis minus c o t to the power of negative 1 end exponent left parenthesis blank t a n blank alpha right parenthesis$ is equal to

maths-General

General

maths-

The minimum value of $left parenthesis s e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent plus left parenthesis blank c o s blank e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent$ is

maths-General

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maths-

If $5 f left parenthesis x right parenthesis plus 3 f left parenthesis fraction numerator 1 over denominator x end fraction right parenthesis equals x plus 2$ andy $equals x f$ (x)then $fraction numerator d y over denominator d x end fraction$ at x=1 is

maths-General

General

maths-

Statement 1:Degree of differential equation $2 x minus 3 y plus 2 equals blank l o g blank left parenthesis fraction numerator d y over denominator d x end fraction 1$ is not defined
Statement 2: In the given D.E, the power of highest order derivative when expressed as a polynomials of derivatives is called degree

maths-General

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maths-

The solution of $e to the power of x y end exponent left parenthesis x y to the power of 2 end exponent d y plus y to the power of 3 end exponent d x right parenthesis plus e to the power of x divided by y end exponent left parenthesis y d x minus x d y right parenthesis equals 0$ is

maths-General

General

maths-

Solution of the equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y to the power of 2 end exponent minus y minus 2 over denominator x to the power of 2 end exponent plus 2 x minus 3 end fraction$ is $minus$ ( where c is arbitrary constant)

maths-General

General

Maths-

Assertion (A) : If (2, 3) and (3,2) are respectively the orthocenters of $capital delta A B C$ and $capital delta D E F$ where D, E, F are mid points of sides of $capital delta A B C$ respectively then the centroid of either triangle is $Error converting from MathML to accessible text.$
Reason (R) :Circumcenters of triangles formed by mid‐points of sides of $capital delta A B C$ and pedal triangle of $capital delta A B C$ are same

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The points of contact of the vertical tangents to $x equals 2 minus 3 blank s i n blank theta comma$ $y equals 3 plus 2 blank c o s blank theta$ are

(2,5), (2,1)
(-1,3), (5,3)
(2,5)(5,3)
(-1,3)(2,1)

The correct answer is: (-1,3), (5,3)

Related Questions to study

The curve $y equals a x to the power of 3 end exponent plus b x to the power of 2 end exponent plus c x plus 5$ touches the $x$ ‐axis at P(-2,0) and cuts the y ‐axis at a point $Q left parenthesis O comma 5 right parenthesis$ where its gradient is 3 Then a+2b+c=-

The curve $y equals a x to the power of 3 end exponent plus b x to the power of 2 end exponent plus c x plus 5$ touches the $x$ ‐axis at P(-2,0) and cuts the y ‐axis at a point $Q left parenthesis O comma 5 right parenthesis$ where its gradient is 3 Then a+2b+c=-

If $open vertical bar table row alpha beta row gamma cell negative alpha end cell end table close vertical bar$ is to be square root of a determinant of the two rowed unit matrix, then $alpha comma beta comma$ $gamma$ should satisfy the relation

If $open vertical bar table row alpha beta row gamma cell negative alpha end cell end table close vertical bar$ is to be square root of a determinant of the two rowed unit matrix, then $alpha comma beta comma$ $gamma$ should satisfy the relation

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Matrix Ais given by $A equals open curly brackets table attributes columnalign left end attributes row 3 11 row 2 8 end table close curly brackets$ then the determinant of $A to the power of 2011 end exponent minus 5 A to the power of 2010 end exponent$ is

The minimum value of $left parenthesis s e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent plus left parenthesis blank c o s blank e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent$ is

The minimum value of $left parenthesis s e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent plus left parenthesis blank c o s blank e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent$ is

If $5 f left parenthesis x right parenthesis plus 3 f left parenthesis fraction numerator 1 over denominator x end fraction right parenthesis equals x plus 2$ andy $equals x f$ (x)then $fraction numerator d y over denominator d x end fraction$ at x=1 is

If $5 f left parenthesis x right parenthesis plus 3 f left parenthesis fraction numerator 1 over denominator x end fraction right parenthesis equals x plus 2$ andy $equals x f$ (x)then $fraction numerator d y over denominator d x end fraction$ at x=1 is

The solution of $e to the power of x y end exponent left parenthesis x y to the power of 2 end exponent d y plus y to the power of 3 end exponent d x right parenthesis plus e to the power of x divided by y end exponent left parenthesis y d x minus x d y right parenthesis equals 0$ is

The solution of $e to the power of x y end exponent left parenthesis x y to the power of 2 end exponent d y plus y to the power of 3 end exponent d x right parenthesis plus e to the power of x divided by y end exponent left parenthesis y d x minus x d y right parenthesis equals 0$ is

Solution of the equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y to the power of 2 end exponent minus y minus 2 over denominator x to the power of 2 end exponent plus 2 x minus 3 end fraction$ is $minus$ ( where c is arbitrary constant)

Solution of the equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y to the power of 2 end exponent minus y minus 2 over denominator x to the power of 2 end exponent plus 2 x minus 3 end fraction$ is $minus$ ( where c is arbitrary constant)

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The points of contact of the vertical tangents to are

(2,5), (2,1) (-1,3), (5,3) (2,5)(5,3) (-1,3)(2,1)

The correct answer is: (-1,3), (5,3)

Related Questions to study

The curve touches the ‐axis at P(-2,0) and cuts the y ‐axis at a point where its gradient is 3 Then a+2b+c=-

The curve touches the ‐axis at P(-2,0) and cuts the y ‐axis at a point where its gradient is 3 Then a+2b+c=-

If is to be square root of a determinant of the two rowed unit matrix, then should satisfy the relation

If is to be square root of a determinant of the two rowed unit matrix, then should satisfy the relation

Matrix Ais given by then the determinant of is

Matrix Ais given by then the determinant of is

If then

If then

If then the number of positive divisors of the number p is

If then the number of positive divisors of the number p is

The number of positive integral solutions of the equation is

The number of positive integral solutions of the equation is

If be real matrices (not necessasrily square) such that forthematrix consider the statements

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If , thenStatement becauseStatement :A.M for

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If , then the value of is equal to

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The minimum value of is

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If andy (x)then at x=1 is

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Statement 1:Degree of differential equation is not definedStatement 2: In the given D.E, the power of highest order derivative when expressed as a polynomials of derivatives is called degree

Statement 1:Degree of differential equation is not definedStatement 2: In the given D.E, the power of highest order derivative when expressed as a polynomials of derivatives is called degree

The solution of is

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Solution of the equation is ( where c is arbitrary constant)

Solution of the equation is ( where c is arbitrary constant)

Assertion (A) : If (2, 3) and (3,2) are respectively the orthocenters of and where D, E, F are mid points of sides of respectively then the centroid of either triangle is Reason (R) :Circumcenters of triangles formed by mid‐points of sides of and pedal triangle of are same

Assertion (A) : If (2, 3) and (3,2) are respectively the orthocenters of and where D, E, F are mid points of sides of respectively then the centroid of either triangle is Reason (R) :Circumcenters of triangles formed by mid‐points of sides of and pedal triangle of are same

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The points of contact of the vertical tangents to $x equals 2 minus 3 blank s i n blank theta comma$ $y equals 3 plus 2 blank c o s blank theta$ are

(2,5), (2,1)
(-1,3), (5,3)
(2,5)(5,3)
(-1,3)(2,1)

The curve $y equals a x to the power of 3 end exponent plus b x to the power of 2 end exponent plus c x plus 5$ touches the $x$ ‐axis at P(-2,0) and cuts the y ‐axis at a point $Q left parenthesis O comma 5 right parenthesis$ where its gradient is 3 Then a+2b+c=-

The curve $y equals a x to the power of 3 end exponent plus b x to the power of 2 end exponent plus c x plus 5$ touches the $x$ ‐axis at P(-2,0) and cuts the y ‐axis at a point $Q left parenthesis O comma 5 right parenthesis$ where its gradient is 3 Then a+2b+c=-

If $open vertical bar table row alpha beta row gamma cell negative alpha end cell end table close vertical bar$ is to be square root of a determinant of the two rowed unit matrix, then $alpha comma beta comma$ $gamma$ should satisfy the relation

If $open vertical bar table row alpha beta row gamma cell negative alpha end cell end table close vertical bar$ is to be square root of a determinant of the two rowed unit matrix, then $alpha comma beta comma$ $gamma$ should satisfy the relation

Matrix Ais given by $A equals open curly brackets table attributes columnalign left end attributes row 3 11 row 2 8 end table close curly brackets$ then the determinant of $A to the power of 2011 end exponent minus 5 A to the power of 2010 end exponent$ is

Matrix Ais given by $A equals open curly brackets table attributes columnalign left end attributes row 3 11 row 2 8 end table close curly brackets$ then the determinant of $A to the power of 2011 end exponent minus 5 A to the power of 2010 end exponent$ is

The minimum value of $left parenthesis s e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent plus left parenthesis blank c o s blank e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent$ is

The minimum value of $left parenthesis s e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent plus left parenthesis blank c o s blank e c to the power of negative 1 end exponent x right parenthesis to the power of 2 end exponent$ is

If $5 f left parenthesis x right parenthesis plus 3 f left parenthesis fraction numerator 1 over denominator x end fraction right parenthesis equals x plus 2$ andy $equals x f$ (x)then $fraction numerator d y over denominator d x end fraction$ at x=1 is

If $5 f left parenthesis x right parenthesis plus 3 f left parenthesis fraction numerator 1 over denominator x end fraction right parenthesis equals x plus 2$ andy $equals x f$ (x)then $fraction numerator d y over denominator d x end fraction$ at x=1 is

The solution of $e to the power of x y end exponent left parenthesis x y to the power of 2 end exponent d y plus y to the power of 3 end exponent d x right parenthesis plus e to the power of x divided by y end exponent left parenthesis y d x minus x d y right parenthesis equals 0$ is

The solution of $e to the power of x y end exponent left parenthesis x y to the power of 2 end exponent d y plus y to the power of 3 end exponent d x right parenthesis plus e to the power of x divided by y end exponent left parenthesis y d x minus x d y right parenthesis equals 0$ is

Solution of the equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y to the power of 2 end exponent minus y minus 2 over denominator x to the power of 2 end exponent plus 2 x minus 3 end fraction$ is $minus$ ( where c is arbitrary constant)

Solution of the equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y to the power of 2 end exponent minus y minus 2 over denominator x to the power of 2 end exponent plus 2 x minus 3 end fraction$ is $minus$ ( where c is arbitrary constant)