Question

If the coordinates if the vertex and focus of a parabola re (-1,1) and (2,3) respectively,then write the equation of its directrix.

Answer 1

The equation of line posses through vertex and focus of a parabola is $\frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}$

$\Rightarrow \frac{3-1}{2-(-1)}=\frac{y-1}{x-(-1)}$ $[\because \text{Focus:(2,3) and vertex :(-1,1)}]$

$\Rightarrow \frac{2}{3}=\frac{y-1}{x+1}$

$\Rightarrow 2x+2=3y-3$

$\Rightarrow 3y-2x-3-2=0$

$\Rightarrow 3y-2x-5=0...\left(i\right)$

$\text{The equation of}\perp \text{line to}$

$3y-2x-5=0$

$2y+3x+\lambda =0...\left(ii\right)$

$Let(x_1,y_1)\text{be the coordinates of the point of intersection of the axis and directrix.}$

$\text{Then (-1,1) is the mid-point of the line segment joining (2,3) and}(x_1,y_1).$

$\therefore \frac{x_1+2}{2}=-1and\frac{y_1+3}{2}=1$

$\Rightarrow x_1+2=-2and{y}_1+3=2$

$\Rightarrow x_1=-4and{y}_1=-1$

$\text{Thus, the directrix meets the axis at (-4,1)}$.

$\therefore \text{The perpendicular line}2y+3x+\lambda =0\text{poses through (-4,-1)}.$

$\therefore 2(-1)+3(-4)+\lambda =0$

$\Rightarrow -2-12+\lambda =0$

$\Rightarrow \lambda =14$

Putting $\lambda =14$ in equation (ii), we get 2y+3x+14=0

Hence,the required equation of directrix is 2y+3x+14=0