Question

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

Answer 1

Hint: Bring constants on RHS and add $ 4{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 $ 4{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 $ 5{k^2} $ from the initial 3 terms of the equation. Take -5 common. Now, using $ - 2hx = - 10x $ and $ - 2ky = 4y $ we will get the values of k and h. After that we will be replacing $ {(x - 5)^2} $ with $ {x^2} - 10x + {h^2} $ also, 100 by . And $ {(y - ( - 2))^2} $ with $ {y^2} + 4y + {k^2} $ also -20 with $ - 5{k^2} $ . Finally diving the whole equation by 80 and converting every term into a square form we will get our answer as $ \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 $ 4{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:-
$ 4{x^2} - 40x - 5{y^2} - 20y = - 160 $
At this point, we will add $ 4{h^2} $ to both the sides of the equation:-
$ 4{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({x^2} - 10x + {h^2}) - 5{y^2} - 20y = - 160 + 4{h^2} $
Now, we can subtract $ 5{k^2} $ from the initial 3 terms of the equation:-
$ 4({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({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} = {x^2} - 2hx + {h^2} $ , we will use the equation $ - 2hx = - 10x $
Solving $ - 2hx = - 10x $ we will get that $ \dfrac{{{{(y - ( - 2))}^2}}}{{{4^2}}} - \dfrac{{{{(x - 5)}^2}}}{{{{(2\sqrt 5 )}^2}}} = 1 $
Now, replacing $ {(x - 5)^2} $ with $ {x^2} - 10x + {h^2} $ also, 100 by $ 4{h^2} $
So,
$ 4{(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} = {y^2} - 2ky + {k^2} $ , we use the equation:-
$ - 2ky = 4y $
This gives $ k = - 2 $
Now, replacing $ {(y - ( - 2))^2} $ with $ {y^2} + 4y + {k^2} $ also -20 with $ - 5{k^2} $
$ 4{(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} = - 80 $
Now, we will divide both the sides by -80 to get:-
$ \dfrac{{{{(y - ( - 2))}^2}}}{{16}} - \dfrac{{{{(x - 5)}^2}}}{{20}} = 1 $
Converting the denominators into squares, we can write it as:-
$ \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.