Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If y=mx+5 is a tangent to x3y3=ax3+by3 at point 1,2, then the value of a is equal to
MathematicsApplication of DerivativesJEE Main
Options:
  • A 95
  • B 165
  • C 94
  • D 187
Solution:
2562 Upvotes Verified Answer
The correct answer is: 165
P1,2 lies on the curve and line, hence
L2=m+5m=-3
Curve8=a+8b
Also, dydx at P1,2 for the curve should be m=-3,
i.e. 3x2y3+x3.3y2y'=3ax2+3by2y'
24+12-3=3a+12b-3
-12=3a-36b
a-12b=-4 & a+8b=8
-20b=-12
b=35 and a=165

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.