wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.
​​​​​​Column 1Column 2Column 3(I) x2+y2=a2(i) my=m2x+a(P) (am2,2am)(II) x2+a2y2=a2(ii) y=mx+am2+1(Q) (mam2+1,am2+1)(III) y2=4ax (iii) y=mx+a2m21(R) (a2ma2m2+1,1a2m2+1)(IV) x2a2y2=a2(iv) y=mx+a2m2+1(S) (a2ma2m21,1a2m21)

The tangent to a suitable conic (Column 1) at (3,12) is found to be 3x+2y=4, then which of the following options is the only CORRECT combination?

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

The correct option is D (II)(iv)(R)
Tangent at (3,12) is found to be 3x+2y=4.
Since, slope of tangent at (3,12) is negative, hence possible curve are (I) and (II) only.
Therefore, equation of curve is
x24+y2=1x2+a2y2=a2.
Hence, the equation of tangent is y=mx+a2m2+1.
And the point of contact is (a2ma2m2+1,1a2m2+1)

The correct combination of curve, tangent and point of contact is given in the table below:
(I)(ii)(Q)(II)(iv)(R)(III)(i)(P)(IV)(iii)(S)

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Line and Ellipse
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon