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3 years ago

Q9: Determine e, k, and identify the type of conic for r= 12/7-cos theta .

Mathematics

2 answers:

kicyunya [14]3 years ago

6 0

Answer:

eccentricity; e = 1/7

k = 12

Conic section; Ellipse

lapo4ka [179]3 years ago

3 0

Answer:

eccentricity; e = 1/7

k = 12

Conic section; Ellipse

Step-by-step explanation:

The first step would be to write the polar equation of the conic section in standard form by multiplying the numerator and denominator by 1/7;

$r=\frac{\frac{12}{7} }{1-\frac{1}{7}cos theta}$

The polar equation of the conic section is now in standard form;

The eccentricity is given by the coefficient of cos theta in which case this would be the value 1/7. Therefore, the eccentricity of this conic section is 1/7.

The eccentricity is clearly between 0 and 1, implying that the conic section is an Ellipse.

The value in the numerator gives the value of k; k = 12

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Tems11 [23]

$\bold{\huge{\underline{ Solution }}}$

<h3>Given :- </h3>

• $\sf{ Polynomial :- ax^{2} + bx + c }$

• The zeroes of the given polynomial are α and β .

<h3>Let's Begin :- </h3>

Here, we have polynomial

$\sf{ = ax^{2} + bx + c }$

We know that, 

Sum of the zeroes of the quadratic polynomial

$\sf{ {\alpha} + {\beta} = {\dfrac{-b}{a}}}$

And 

Product of zeroes

$\sf{ {\alpha}{\beta} = {\dfrac{c}{a}}}$

Now, we have to find the polynomials having zeroes :-

$\sf{ {\dfrac{{\alpha} + 1 }{{\beta}}} ,{\dfrac{{\beta} + 1 }{{\alpha}}}}$

Therefore ,

Sum of the zeroes

$\sf{ ( {\alpha} + {\dfrac{1 }{{\beta}}} )+( {\beta}+{\dfrac{1 }{{\alpha}}})}$

$\sf{ ( {\alpha} + {\beta}) + ( {\dfrac{1}{{\beta}}} +{\dfrac{1 }{{\alpha}}})}$

$\sf{( {\dfrac{ -b}{a}} ) + {\dfrac{{\alpha}+{\beta}}{{\alpha}{\beta}}}}$

$\sf{( {\dfrac{ -b}{a}} ) + {\dfrac{-b/a}{c/a}}}$

$\sf{ {\dfrac{ -b}{a}} + {\dfrac{-b}{c}}}$

$\bold{{\dfrac{ -bc - ab}{ac}}}$

Thus, The sum of the zeroes of the quadratic polynomial are -bc - ab/ac

Product of zeroes

$\sf{ ( {\alpha} + {\dfrac{1 }{{\beta}}} ){\times}( {\beta}+{\dfrac{1 }{{\alpha}}})}$

$\sf{ {\alpha}{\beta} + 1 + 1 + {\dfrac{1}{{\alpha}{\beta}}}}$

$\sf{ {\alpha}{\beta} + 2 + {\dfrac{1}{{\alpha}{\beta}}}}$

$\bold{ {\dfrac{c}{a}} + 2 + {\dfrac{ a}{c}}}$

Hence, The product of the zeroes are c/a + a/c + 2 .

We know that, 

For any quadratic equation

$\sf{ x^{2} + ( sum\: of \:zeroes )x + product\:of\: zeroes }$

$\bold{ x^{2} + ( {\dfrac{ -bc - ab}{ac}} )x + {\dfrac{c}{a}} + 2 + {\dfrac{ a}{c}}}$

Hence, The polynomial is x² + (-bc-ab/c)x + c/a + a/c + 2 .

<h3>Some basic information :-</h3>

• Polynomial is algebraic expression which contains coffiecients are variables.

• There are different types of polynomial like linear polynomial , quadratic polynomial , cubic polynomial etc.

• Quadratic polynomials are those polynomials which having highest power of degree as 2 .

• The general form of quadratic equation is ax² + bx + c.

• The quadratic equation can be solved by factorization method, quadratic formula or completing square method.

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the length of a rectangle is 7 inches more than its width, if the perimiter of the rectangle is 66 inches, find its dimensions.

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Perimeter of rectangle=66
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width of rectangle=w
P of a rect.= 2(length)+ 2(width)
66= 2L+2w

if the length is 7in more than the width, then
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Now we will substitute 7+w in for L. Here is our new equation:

66=2(7+w) + 2w

Solve for w

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66=14+4w
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I hooe this is explained well enough

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