give me a complete and managed answer.​

Answer:

Solution given:

The given equation of a line is

ax²+2hxy+by²=0

Let y=mx be any one line of

ax²+2hxy+by²=0

Let the perpendicular line of

y=mx is

x+my=0

According to the question the line x+my=0 is one line of Ax²+2Hxy +By²=0

Substituting value of y

Ax²+2Hx [tex] \frac{ - x}{m} [/tex]+[tex]B \frac{x²}{m²} [/tex]=0

Ax²-[tex] \frac{ 2H}{m}x² [/tex]+[tex]B \frac{x²}{m²} [/tex]=0

Taking LCM

we get

Ax²m²-2mHx²+Bx²=0

x²[Am²-2Hm+B]=0..............[1]

Again.

ax²+2hx(mx)+B(mx)²=0

ax²+2hmx²+Bm²x²=0

x²[a+2hm+Bm²]=0

bm²+2hn+a=0.....................[2]

Taking coefficient of equation 1 &2.

equation 1. b 2h a b 2h

equation 2.A. -2h B A -2H

Doing criss cross multiplication

ignore first coefficient and repeat first and second

again

lines are perpendicular so

[tex] \frac{m²}{2hB} [/tex]=[tex] \frac{m}{aA-bB} [/tex]=[tex]\frac{1}{-2Hb-2hA} [/tex]

Taking 1st & 2nd ratio,we get,

m=[tex] \frac{2hB+2Ha}{aA-bB} [/tex]....[*]

Taking 3rd & 2nd ratio,we get,

m=[tex] \frac{aA-bB}{-2Hb-2hA} [/tex] ....[#]

Again

Equating equation * &# we get;

[tex] \frac{2hB+2Ha}{aA-bB} [/tex]=[tex] \frac{aA-bB}{-2Hb-2hA} [/tex]

(aA-bB)²=-4(hB+Ha)(Hb+HA)

(aA-bB)²+4(hB+Ha)(Hb+HA)=0 is a required condition.

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please help me thank you

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Your answer

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(NO LINKS)
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Hot Sauce (oz) 0 5 6 60
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The optimal number of servings of Wimpy is
The optimal number of servings of Dial 911 is
The optimal number of servings of Fire Bowl is
The maximum possible profit is
Is there any leftover of ingredients as a result of this optimal solution? if yes, enter the amount. Otherwise, just type 0.
Amount of beef leftover
Amount of onion leftover
Amount of special sauce leftover
Amount of hot sauce leftover

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ANSWER ASAP DONT SEND A FILE WHATS THE RULE TRANSFORMATION FOR THIS DILATION ???