##
The equation of tangent to the curve y (1 + x
^{
2
}
) = 2 – x, where it crosses x-axis is:

## (A) x + 5y = 2 (B) x – 5y = 2

## (C) 5x – y = 2 (D) 5x + y = 2

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Chapter 6 Class 12 Application of Derivatives

Serial order wise

Last updated at April 16, 2024 by Teachoo

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)