0 votes
328 views
in Uncategorized by (98.9k points)
edited

The equation of the common tangent to the curves y2 = 16x and xy = - 4 is

  • (a) x-y+4=0
  • (b) x+y+4=0
  • (c) x-2y+16=0
  • (d) 2x-y+2=0

1 Answer

0 votes
by (98.9k points)
selected by
 
Best answer

Correct option is (d) y=x+2 

Explaination::

tangents to the curve y2=16x is y=mx+4/m , so it must satisfy xy=-4 

\(x({mx+\frac{4}{m}})=-4\)

\(mx^2+\frac{4}{m}x+4=0 ,\)

since it has equal roots , therefore D=0

\(\frac{16}{m^2}-16m=0\)

\(m^3=1 \)

m=1

therefore , the equation of commom tangent is  y=x+4 i.e. x-y+4=0

Related questions

Doubtly is an online community for engineering students, offering:

  • Free viva questions PDFs
  • Previous year question papers (PYQs)
  • Academic doubt solutions
  • Expert-guided solutions

Get the pro version for free by logging in!

5.7k questions

5.1k answers

108 comments

561 users

...