Question

Find the eccentricity of the ellipse 3x² + 4y² = 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

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

Differentiate both sides with respect to $x$

$$6x+8yy'=0$$

$$y'=-\dfrac{3x}{4y}$$

Point $(1, 3/2)$

$$y'(1, \dfrac{3}{2})=-\dfrac{3(1)}{4(\dfrac{3}{2})}=-\dfrac{1}{2}$$

The equation of the tangent in point-slope form

$$y-\dfrac{3}{2}=-\dfrac{1}{2}(x-1)$$

The equation of the tangent in slope-intercept form

$$y=-\dfrac{1}{2}x+2$$

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

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

$$a^2=4, b^2=3, c=\sqrt{a^2-b^2}=\sqrt{4-3}=1$$

The eccentricity of the ellipse

$$e=\dfrac{c}{a}=\dfrac{1}{2}$$

Foci: $(\pm c, 0)$

$$S(-1, 0), S'(1, 0)$$

$$x=0: y(0)=-\dfrac{1}{2}(0)+2=2$$

$$G(0,2)$$

Triangle $SGS'$

$$Area= \dfrac{a}{b}(2)(1-(-1))=2 (units^2)$$

The area of triangle SS′G is 2 square units.