Max. and Min. value of r

For convenience, I'm writing the equation as 5r² + 12r.cosθ + 7 = 0.

By using quadratic formula,
r =-(12cosθ) ± √144cos²Î¸ - 4.5.72x5

=-12cosθ ± 2√36cos²Î¸ - 3510

=-6cosθ ± √36cos²Î¸ - 355

Now since the maximum value of cosθ is (+1),
putting cosθ=+1, we get

r = -6(1) ± √36.(1)² - 355 = - 75 or -1.

Since minimum value of cosθ is (-1), putting cosθ = -1, we get,
r = -6(-1) ± √36(-1)² - 355 = 75 or 1

But since r≥0,
Maximum value of r is 75..
Minimum value of r is 1.. [Ans]

@ Abhisekh, how do you guarantee that the numerator is increasing in cosθ ?

Set -6cos\alpha \pm \sqrt{36cos^2\alpha-35}=k

Squaring and bringing the terms to 1 side,we get k^2+12kcos\alpha+35=0\Rightarrow -1\le \frac{-(k^2+35)}{12k}\le 1.

That helps, I guess.

how do you guarantee that the numerator is increasing in cosθ ?

What if it doesn't? Sorry But I couldn't get it.

ya soumik
e1 i cudnt understand what u wanna point out??

he means, wats d guarantee that numerator also increases as cosθ increases

5r² +12rcosθ + 7 = 0 ......(1)

differentiate wrt θ
→ drdθ = 12rsinθ
10r+12cosθ

for r max or min , dr/dθ = 0

hence r= 0 or sinθ = 0

but r=0 is not permissible bcz of eqn 1 ,

so sinθ= 0

so cosθ = ±1, put this in eqn 1

5r² - 12r + 7 = 0 , i.e r = 1 or 7/5

5r² + 12r + 7 = 0, i.r r = -1 or - 7/5
hence r_min = 1 , r_max = 7/5 as r≥ 0

@ qwerty:

Can u please specify what was wrong with my method?

I never said that numerator would increase as cosθ increases....in fact, the maximum and minimum values are coming only for the minimum value of cosθ
i.e. -1....

(and the answers also match.)

:-|

how did u came to kno that the value of r that u obtained by putting minimum value of cosθ is the minimum/maximum possible value of r ?

I went on with the assumption that the extreme values of r would come with the extreme values of cosθ , irrespective of whether the value of cosθ is its maximum or minimum....

But anyways, MISTAKE ACCEPTED !

Thanks everyone for clearing my misconception ! :)