CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

The line 3x+4y=7 touches the ellipse 3x2+4y2=1 at the point

A
(7,7)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(17,17)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(17,17)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(7,7)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (17,17)
Let the given line touches the ellipse at point P(θ)=(cosθ3,sinθ2)

The equation of the tangent at P is
3xcosθ+2ysinθ=1(i)

Comparing equation (i) with the given equation of the line 3x+4y=7,
we get
cosθ3=sinθ2=17

Point of contact is (17,17)

Alternate Solution:
For the given ellipse a2=1/3,b2=1/4
Given tangent is
3x+4y=7y=34x+74
m=34,c=74
Using slope form the point of contact will be
(a2mc,b2c)(17,17)

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image