Struggling with this Conics Question (1 Viewer)

Sy123 · Mar 20, 2014

$\\ $Consider the general line through$ \ \ P(\alpha, \beta) \ \ $of the form$ \\ y=mx+c \ \ $intersecting the ellipse at$ \ \ (x_1,y_1) \ ,\ (x_2,y_2) \\ \\ $We find that solving for the intersections (of which$ \ P \ $is midpoint of)$ \\ \\ x^2(b^2+m^2a^2) + 2mca^2x + a^2c^2 - a^2b^2 = 0 \\ \\ x_1 + x_2 = \frac{-2mca^2}{b^2+a^2m^2} = 2\alpha \\ \\ $Therefore$ \ \ \alpha = \frac{-mca^2}{b^2+a^2m^2} \\ \\ $Doing the same for$ \ y-$values$ \\ x= \frac{y-c}{m} \\ \\ y^2(b^2+a^2m^2) - 2b^2cy + b^2c^2 - a^2b^2m^2 = 0 \\ \\ $Therefore$ \ \ \beta = \frac{b^2c}{b^2+a^2m^2}$

$\therefore \ \frac{\beta}{\alpha} = \frac{b^2c}{b^2+a^2m^2} \cdot \frac{b^2+a^2m^2}{-mca^2} \Rightarrow \ m = - \frac{\alpha b^2}{\beta a^2}$

$\\ \text{Since} \ P(\alpha, \beta) \ \text{lies on} \ y=mx+c \ \beta -m\alpha = c \\ \\ c = \frac{a^2\beta^2 + b^2 \alpha^2}{a^2\beta}$

$\\ $Substitute$ \ m,c \ $back into equation and simplify to yield the desired result$$

Struggling with this Conics Question (1 Viewer)

Cleavage

Clarence

Cleavage

Clarence

Sy123

This too shall pass

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