Question

if an ellipse passing through (3,1) having foci (+or - 4, 0) then find 1)the length of major axis. 2)the standard equation of ellipse. 3)the eccentricity and length of the latus rectum​

Answer 1

Answer:

1.Length of major axis$=6\sqrt{2}$

2.Equation of ellipse is

$\frac{x^2}{18}+\frac{y^2}{2}=1$

3.Eccentricity$=\frac{2\sqrt{2}}{3}$

4.Length of latusrectum$=\frac{2\sqrt{2}}{3}$

Step-by-step explanation:

$Given:\\\\F_1(4,0)\:and\:F_2(-4,0)\\\\But \\\\F_1F_2=2ae\\8 =2ae\\ae = 4\\\\\\b^2=a^2(1-e^2)=a^2-(ae)^2=a^2-16$

Centre C is$(\frac{4+(-4)}{2},\frac{0+0}{0})$

Centre C is (0,0)

since the major axis is along x-axis, the equation of ellipse is of the form

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\\\\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=1$

But it passes through (3,1)

$\frac{3^2}{a^2}+\frac{1^2}{a^2-16}=1\\\\\frac{9}{a^2}+\frac{1}{a^2-16}=1\\\\\frac{9(a^2-16)+a^2}{a^2(a^2-16)}=1\\\\9a^2-144+a^2=a^4-16a^2\\\\a^4-26a^2+144=0\\\\(a^2-18)(a^2-8)=0\\\\a^2=18\:or\:8when \:a^2=8,\: b^2=-8,$

It is not possible

$when\:a^2=18,\:b^2=2$

The equation of ellipse is

$\frac{x^2}{18}+\frac{y^2}{2}=1$

$Eccentricity\\\\=\sqrt{1-\frac{b^2}{a^2}}\\\\=\sqrt{1-\frac{2}{18}}\\\\=\sqrt{1-\frac{1}{9}}\\\\=\sqrt{\frac{8}{9}}\\\\=\frac{2\sqrt{2}}{3}$

$a^2=18\\\\a=3\sqrt{2}$

Length of the major axis $= 2a=6\sqrt{2}$

$Length \:of \:latusrectum=\frac{2b^2}{a}\\\\=\frac{2(2)}{3\sqrt{2}}\\\\=\frac{2\sqrt2}{3}$

Answer 2

length of major axis=6√2

standard equation of ellipse : (x²/18)+(y²/2)=1

eccentricity=⅔√2

length of latus rectum=⅔√2