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a^2= 5^2= 25
b^2=3^2=9
Vậy phương trình chính tắc của elip là:
x^2/25+y^2/9=1
–Thay a² và b² vào dạng tổng quát, phương trình chính tắc của elip là:
x²/25 + y²/9 = 1
a: Tọa độ trung điểm I của AB là:
\(\begin{cases}x_{I}=\frac{x_{A}+x_{B}}{2}=\frac{1-2}{2}=-\frac12\\ y_{I}=\frac{y_{A}+y_{B}}{2}=\frac{2+3}{2}=\frac52\end{cases}\)
=>I(-0,5;2,5)
A(1;2); B(-2;3)
=>\(\overrightarrow{AB}=\left(-2-1;3-2\right)=\left(-3;1\right)\)
(d) là đường trung trực của AB
=>(d)⊥AB tại trung điểm I của AB
=>(d) nhận \(\overrightarrow{AB}=\left(-3;1\right)\) làm vecto pháp tuyến và đi qua I(-0,5;2;5)
Phương trình (d) là:
-3(x+0,5)+1(y-2,5)=0
=>-3x-1,5+y-2,5=0
=>-3x+y-4=0
b: (Δ1)//(Δ2)
=>(Δ1): x-y+c=0
Thay x=1 và y=2 vào (Δ1), ta được:
1-2+c=0
=>c-1=0
=>c=1
=>(Δ1): x-y+1=0
c: A(1;2); B(-2;3)
=>\(AB=\sqrt{\left(-2-1\right)^2+\left(3-2\right)^2}=\sqrt{3^2+1^2}=\sqrt{10}\)
Phương trình đường tròn (C) là:
\(\left(x-1\right)^2+\left(y-2\right)^2=AB^2=10\)
d:
x-y+3=0
=>y=x+3
M thuộc (Δ2) nên M(x;x+3)
M(x;x+3); A(1;2)
MA=2
=>\(MA^2=4\)
=>\(\left(x-1\right)^2+\left(x+3-2\right)^2=4\)
=>\(\left(x-1\right)^2+\left(x+1\right)^2=4\)
=>\(x^2-2x+1+x^2+2x+1=4\)
=>\(2x^2=2\)
=>\(x^2=1\)
=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)
Khi x=1 thì M(1;4)
Khi x=-1 thì M(-1;2)
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Phương trình đường tròn (C):x^2+y^2=8
Phương trình elip(E):x^2/16+y^2/8=1
a)Phương trình đường tròn (C):
x^2 + y^2 = 8.
b)Phương trình chính tắc của elip (E):
x^2/16 + y^2/8 = 1.