Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone explain - Limit , continuity and differentiability - JEE Main-14

Equation of tangent at (3, 4) being on x^2 + y^2 -4x -6y + 11 = 0 is

  • Option 1)

    x + y = 7

  • Option 2)

    x -y = -1

  • Option 3)

    2x + y = 10

  • Option 4)

    3x -y = 5

 

Answers (1)

best_answer

As we have learnt,

 

Equation of tangent -

Let  (x_{\circ }\:y_{\circ })  is the point on the curve y = f(x) then direct we can write as 

x^{2}<xx_{\circ },\:\:\:\:\: x=\frac{x+x_{\circ }}{2}


y^{2}<yy_{\circ },\:\:\:\:\: y=\frac{y+y_{\circ }}{2}

for\:\:\:x^{2}+y^{2}=2,\;\:equation\:of\:tangent\:at\:(1,\:1)\:is\:x\times 1+y\times 1=2

\therefore \:x+y=2

-

 

 

xx_1 + yy_1 -2(x + x_1) - 3(y + y_1) + 11\Rightarrow will be equation of tangent at (x_1, y_1).

So equation becomes x(3) + y(4) -2(x + 3) - 3(y + 4) + 11 = 0 \Rightarrow x + y = 7


Option 1)

x + y = 7

Option 2)

x -y = -1

Option 3)

2x + y = 10

Option 4)

3x -y = 5

Posted by

Himanshu

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question