Answer to Question #104571 in Geometry for Max

Question

Find the eccentricity of the ellipse 3x
2 + 4y
2 = 12 and the equation of the tangent to
the ellipse at the point (1, 3/2). If this tangent meets the y-axis at the point G, and
S and S
′ are the foci of the ellipse, find the area of triangle SS′G.

Answer 1

$3x^2+4y^2=12$

divided by 12 on both sides

${3 \above{2pt} 12}x^2+{4 \above{2pt} 12}y^2=1$

${x^2 \above{2pt} 4}+{y^2 \above{2pt} 3}=1$

The eccentricity is a measure of how much the ellipse deviates from the circle.For an Ellipse with major axis parallel to the x-axis, the eccentricity is ${ \sqrt{a^2-b^2}\above{2pt} a}$

$=$ ${ \sqrt{a^2-b^2}\above{2pt} a}$

Calculate Ellipse properties

$3x^2+4y^2=12$ : Ellipse with center $(h,k)=(0,0)$ , semi-major axis $a=2$ ,semi -minor axis

$b=\sqrt{3}$

$3x^2+4y^2=12$

Ellipse Standard Equation:

${(x-h)^2 \above{2pt} a^2}+{(y-k)^2 \above{2pt} b^2}=1$ is the Ellipse standard equation with center (h,k) and a,b are the semi-major and semi-minor axes .

Rewrite $3x^2+4y^2=12$ in the form of standard ellipse equation .

$3x^2+4y^2=12$

divided by 12 on both sides

${3 \above{2pt} 12}x^2+{4 \above{2pt} 12}y^2=1$

${x^2 \above{2pt} 4}+{y^2 \above{2pt} 3}=1$

Rewrite in standard form

${(x-0)^2 \above{2pt} 2^2}+{(y-0)^2 \above{2pt} (\sqrt{3})^2}=1$

Therefore Ellipse properties are :

$(h,k)=(0,0), a=2 ,b=\sqrt{3}$

$a>b$ therefore a is semi-major axis and b is semi-minor axis.

Ellipse with center $(h,k)=(0,0)$ ,semi-major axis $a=2$ , semi-minor axis $b=\sqrt{3},$

${ \sqrt{2^2-(\sqrt{3})^2}\above{2pt} 2}$

$={1 \above{2pt} 2}$

the equation of the tangent to

the ellipse at the point (1, 3/2). If this tangent meets the y-axis at the point G, and

S and S

Equation tangent will be

$y=mx+\sqrt{a^2m^2+b^2}$

$y=mx+\sqrt{4m^2+(\sqrt{3})^2}$

As this line passes through (1,3/2 )

Therefore ${3 \above{2pt} 2}-m=$ $\sqrt{4m^2+(\sqrt{3})^2}$

$\implies$ ${9 \above{2pt} 4}+m^2-3m=4m^2+(\sqrt{3})^2$

$\implies 9+4m^2-12m=16m^2+4(\sqrt{3})^2$

$\implies 9+4m^2-12m-16m^2-12=0$

$\implies -12m^2-12m-3=0$

by quadratic formula

$m=-{1 \above{2pt} 2}$

$y=-{1\above{2pt} 2}x+2$

Area of triangle SS'G=${1 \above{2pt} 2}\sqrt{({1 \above{2pt} 2})^2+(1)^2}$

$={1 \above{2pt} 2}$ $\sqrt{{5 \above{2pt} 4}}$

$={1 \above{2pt} 4}\sqrt{5}$