Question
In Exercises $55-72$ , find the vertex, focus, and directrix of the parabola and sketch its graph. $$y^{2}+x+y=0$$
Step 1
To do this, we complete the square for the $y$ terms. We have $y^{2}+y$ which can be written as $(y+1/2)^{2} - (1/2)^{2}$. So, the equation becomes $(y+1/2)^{2} - (1/4) + x = 0$. Show more…
Show all steps
Your feedback will help us improve your experience
Suman Saurav Thakur and 93 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
In Exercises $55-72$ , find the vertex, focus, and directrix of the parabola and sketch its graph. $$ x+y^{2}=0 $$
Topics in Analytic Geometry
Circles and Parabolas
In Exercises $55-72$ , find the vertex, focus, and directrix of the parabola and sketch its graph. $$ x^{2}+8 y=0 $$
In Exercises $55-72$ , find the vertex, focus, and directrix of the parabola and sketch its graph. $$ x^{2}+4 x+6 y-2=0 $$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD