Question 12
The equation of tangent to the curve y (1 + x2) = 2 β x, where it crosses x-axis is:
(A) x + 5y = 2 (B) x β 5y = 2
(C) 5x β y = 2 (D) 5x + y = 2
First, let us find the point at which curve cuts π₯βaxis
Since π=π on π₯βaxis
β΄ Point =(π,π)
For πβ coordinate
Putting π=π in
π¦(1+π₯^2 )=2βπ₯
0(1+π₯^2 )=2βπ₯
0=2βπ₯
π=π
β΄ Point =(π,π)
Now, to find equation of tangent we find slope of tangent at (π, π)
Finding Slope of tangent
Given curve
π¦(1+π₯^2 )=2βπ₯
Differentiating w.r.t x
ππ¦/ππ₯ (1+π₯^2 )+π¦ Γ (0+2π₯)=0β1
ππ¦/ππ₯ (1+π₯^2 )+π¦(2π₯)=β1
ππ¦/ππ₯ (1+π₯^2 )+2π₯π¦=β1
ππ¦/ππ₯ (1+π₯^2 )=β1β2π₯π¦
π π/π π=(βπ β πππ" " )/((π + π^π ) )Since we need Slope at (2, 0)
Putting x = 2, y = 0 in ππ¦/ππ₯
π π/π π =(β1 β (2 Γ 2 Γ 0))/(1 + 2^2 )
=(β1 β 0)/(1 + 4)
=(βπ)/π
Finding equation of tangent
Equation of line at (π₯1 , π¦1) & having Slope m is
π¦βπ¦1=π(π₯βπ₯1)
β΄ Equation of tangent at (2, 0) & Slope (β1)/5 is
(πβπ)=(βπ)/π (πβπ)
π¦=(βπ)/π (π₯β2)
5π¦=β1(π₯β2)
5π¦=βπ₯+2
5π¦+π₯=2
π+ππ=π
Hence, equation of tangent is π+ππ=π
So, the correct answer is (A)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!