4.3-bo'lim : Ellipslar

1-7 muammolar uchun ellipsni chizing.

1. \( \displaystyle \frac{{{{\left( {x + 5} \right)}^2}}}{4} + \frac{{{{\left( {y - 2} \right)}^2}}}{9} = 1\)
2. \( \displaystyle {\left( {x - 4} \right)^2} + \frac{{{y^2}}}{{16}} = 1\)
3. \( \displaystyle \frac{{{{\left( {x + 1} \right)}^2}}}{{25}} + \frac{{{{\left( {y + 6} \right)}^2}}}{4} = 1\)
4. \( \displaystyle \frac{{{{\left( {x - 3} \right)}^2}}}{5} + \frac{{{{\left( {y + 1} \right)}^2}}}{{12}} = 1\)
5. \(9{\left( {x - 2} \right)^2} + 4{\left( {y - 3} \right)^2} = 1\)
6. \( \displaystyle \frac{{{{\left( {x - 3} \right)}^2}}}{9} + 2{\left( {y + 4} \right)^2} = 1\)
7. \( \displaystyle \frac{{{{\left( {x - 4} \right)}^2}}}{9} + \frac{{{{\left( {y - 1} \right)}^2}}}{9} = 1\)

8 - 10 masalalar uchun tenglamaning \(x\) va \(y\) qismlaridagi kvadratni to'ldiring va tenglamani ellips tenglamasining standart shakliga yozing.

1. \(4{x^2} - 16x + {y^2} + 2y + 13 = 0\)
2. \({x^2} + 6x + 4{y^2} + 16y + 9 = 0\)
3. \(5{x^2} + 10x + 3{y^2} - 6y - 7 = 0\)