Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

Find the equation of parabola whose vertex is (4,2) and focus is (6,2)?

$\begin{array}{1 1}(a)\;(y-3)^2=8(k-4)\\(b)\;(y-2)^2=8(k-4)\\(c)\;(y-2)^2=4(k-4)\\(d)\;(y-2)^2=6(k-4)\end{array}$

Can you answer this question?

1 Answer

0 votes
Let vertex P(4,2) and focus Q is (6,2)
Slope of PQ=0
Hence axis of parabola parallel to x-axis
The equation is of the form
(h,k) is vertex
Hence equation of parabola is
Hence (b) is the correct answer.
answered Feb 6, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App