Question 10
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.

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