CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Equation of the parabola whose axis is y=x, distance from origin to vertex is 2 and distance from origin to focus is 22, is (Focus and vertex lie in 1st quadrant) :


A

(x+y)2=2(x+y2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

(xy)2=8(x+y2)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

(xy)2=4(x+y2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

(x+y)2=4(x+y2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

(xy)2=8(x+y2)


The equation of axis of the parabolain parametric form is x0cos 45o=y0sin 45o=2 for A,22 for S
A is (1, 1) and S is (2, 2) and foot of directrix be z, then A is mid-point of SZ

x+22=1,y+22=1,z is (0,0).
Equation of directrix is y0=1(x0) or x+y=0
By definition if P(x,y) be any point on the parabola then SP = PM
or (x2)2+(y2)2=[x+y2]2
2[x2+y24x4y+8]=(x+y)2 or x2+y22xy=8(x+y2)


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon