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Question: How do you find the standard form of \( 4{x^2} - 5{y^2} - 40x - 20y + 160 = 0 \) ?...

How do you find the standard form of 4x2−5y2−40x−20y+160=04{x^2} - 5{y^2} - 40x - 20y + 160 = 0 ?

Explanation

Solution

Bring constants on RHS and add 4h24{h^2} to both sides to start off with.
Completing this problem will require steps to be done carefully. First subtract 160 from both the sides of the equation. After that add 4h24{h^2} to both the sides of the equation. After that simplify the equation and take 4 common from a part of the equation. Now, subtract 5k25{k^2} from the initial 3 terms of the equation. Take -5 common. Now, using −2hx=−10x- 2hx = - 10x and −2ky=4y- 2ky = 4y we will get the values of k and h. After that we will be replacing (x−5)2{(x - 5)^2} with x2−10x+h2{x^2} - 10x + {h^2} also, 100 by . And (y−(−2))2{(y - ( - 2))^2} with y2+4y+k2{y^2} + 4y + {k^2} also -20 with −5k2- 5{k^2} . Finally diving the whole equation by 80 and converting every term into a square form we will get our answer as (y−(−2))242−(x−5)2(25)2=1\dfrac{{{{(y - ( - 2))}^2}}}{{{4^2}}} - \dfrac{{{{(x - 5)}^2}}}{{{{(2\sqrt 5 )}^2}}} = 1 .

Complete step by step solution:
The given question we have is to find the standard form of 4x2−5y2−40x−20y+160=04{x^2} - 5{y^2} - 40x - 20y + 160 = 0
The very first step, that we will be doing to solve this problem is subtract 160 from both the sides:-
4x2−40x−5y2−20y=−1604{x^2} - 40x - 5{y^2} - 20y = - 160
At this point, we will add 4h24{h^2} to both the sides of the equation:-
4x2−40x+4h2−5y2−20y=−160+4h24{x^2} - 40x + 4{h^2} - 5{y^2} - 20y = - 160 + 4{h^2}
Now, we will simplify the equation and take 4 common from a part of the equation:-
4(x2−10x+h2)−5y2−20y=−160+4h24({x^2} - 10x + {h^2}) - 5{y^2} - 20y = - 160 + 4{h^2}
Now, we can subtract 5k25{k^2} from the initial 3 terms of the equation:-
4(x2−10x+h2)−5y2−20y−5k2=−160+4h2−5k24({x^2} - 10x + {h^2}) - 5{y^2} - 20y - 5{k^2} = - 160 + 4{h^2} - 5{k^2}
Now, taking -5 common we will get:-
4(x2−10x+h2)−5(y2+4y+k2)=−160+4h2−5k24({x^2} - 10x + {h^2}) - 5({y^2} + 4y + {k^2}) = - 160 + 4{h^2} - 5{k^2}
Since, we are trying to make it a perfect square of the form (x−h)2=x2−2hx+h2{(x - h)^2} = {x^2} - 2hx + {h^2} , we will use the equation −2hx=−10x- 2hx = - 10x
Solving −2hx=−10x- 2hx = - 10x we will get that (y−(−2))242−(x−5)2(25)2=1\dfrac{{{{(y - ( - 2))}^2}}}{{{4^2}}} - \dfrac{{{{(x - 5)}^2}}}{{{{(2\sqrt 5 )}^2}}} = 1
Now, replacing (x−5)2{(x - 5)^2} with x2−10x+h2{x^2} - 10x + {h^2} also, 100 by 4h24{h^2}
So,
4(x−5)2−5(y2+4y+k2)=−160+100−5k24{(x - 5)^2} - 5({y^2} + 4y + {k^2}) = - 160 + 100 - 5{k^2}
Again, we are trying to make a square of the form (y−k)2=y2−2ky+k2{(y - k)^2} = {y^2} - 2ky + {k^2} , we use the equation:-
−2ky=4y- 2ky = 4y
This gives k=−2k = - 2
Now, replacing (y−(−2))2{(y - ( - 2))^2} with y2+4y+k2{y^2} + 4y + {k^2} also -20 with −5k2- 5{k^2}
4(x−5)2−5(y−(−2))2=−160+100−204{(x - 5)^2} - 5{(y - ( - 2))^2} = - 160 + 100 - 20
Solving all the terms, we will get:-
4(x−5)2−5(y−(−2))2=−804{(x - 5)^2} - 5{(y - ( - 2))^2} = - 80
Now, we will divide both the sides by -80 to get:-
(y−(−2))216−(x−5)220=1\dfrac{{{{(y - ( - 2))}^2}}}{{16}} - \dfrac{{{{(x - 5)}^2}}}{{20}} = 1
Converting the denominators into squares, we can write it as:-
(y−(−2))242−(x−5)2(25)2=1\dfrac{{{{(y - ( - 2))}^2}}}{{{4^2}}} - \dfrac{{{{(x - 5)}^2}}}{{{{(2\sqrt 5 )}^2}}} = 1

Note: Forming equations to the standard form, often requires the need to carry multiple complex multiplications of variables and numbers. Therefore, whenever you are doing any problem like this, you must pay a lot of attention while doing basic arithmetic operations otherwise you will do the whole problem wrong.