Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
For any point on the graph of a parabola, the distance to the focus and the distance to the directrix are equal.
The distance to the focus:
[tex](x,y) \\ F(0,1) \\ \\ d_1=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(0-x)^2+(1-y)^2}=\\=\sqrt{(-x)^2+(1-y)^2}=\sqrt{x^2+1-2y+y^2}[/tex]
The distance to the directrix:
The directrix is a horizontal line, so the distance is the absolute value of the difference of the y-coordinates of the points.
[tex](x,y) \\ y=-1 \hbox{ so any point: } (m,-1) \\ \\ d_2=|y-(-1)|=|y+1|[/tex]
The distances are equal:
[tex]d_1=d_2 \\ \sqrt{x^2+1-2y+y^2}=|y+1| \\ (\sqrt{(x^2+1-2y+y^2})^2=|y+1|^2 \\ x^2+1-2y+y^2=(y+1)^2 \\ x^2+1-2y+y^2=y^2+2y+1 \\ x^2+1-1=y^2-y^2+2y+2y \\ x^2=4y \\ \boxed{y=\frac{1}{4}x^2}[/tex]
The distance to the focus:
[tex](x,y) \\ F(0,1) \\ \\ d_1=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(0-x)^2+(1-y)^2}=\\=\sqrt{(-x)^2+(1-y)^2}=\sqrt{x^2+1-2y+y^2}[/tex]
The distance to the directrix:
The directrix is a horizontal line, so the distance is the absolute value of the difference of the y-coordinates of the points.
[tex](x,y) \\ y=-1 \hbox{ so any point: } (m,-1) \\ \\ d_2=|y-(-1)|=|y+1|[/tex]
The distances are equal:
[tex]d_1=d_2 \\ \sqrt{x^2+1-2y+y^2}=|y+1| \\ (\sqrt{(x^2+1-2y+y^2})^2=|y+1|^2 \\ x^2+1-2y+y^2=(y+1)^2 \\ x^2+1-2y+y^2=y^2+2y+1 \\ x^2+1-1=y^2-y^2+2y+2y \\ x^2=4y \\ \boxed{y=\frac{1}{4}x^2}[/tex]
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.