Question 19 - Finding point when tangent is parallel/ perpendicular - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Finding point when tangent is parallel/ perpendicular
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 19 Deleted for CBSE Board 2024 Exams You are here
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Question 16 Deleted for CBSE Board 2024 Exams
Question 26 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 25 Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 23 Important Deleted for CBSE Board 2024 Exams
Question 17 Deleted for CBSE Board 2024 Exams
Question 18 Important Deleted for CBSE Board 2024 Exams
Finding point when tangent is parallel/ perpendicular
Last updated at April 16, 2024 by Teachoo
Question 19 Find the points on the curve π₯^2+π¦^2 β2π₯ β3=0 at which the tangents are parallel to the π₯βππ₯ππ Given that Tangent is parallel to the π₯βππ₯ππ β΄ Slope of tangent = Slope of π₯βππ₯ππ We know that Slope of tangent is ππ¦/ππ₯ Finding π π/π π π₯^2+π¦^2 β2π₯ β3=0 Given that Tangent is parallel to the π₯βππ₯ππ β΄ Slope of tangent = Slope of π₯βππ₯ππ We know that Slope of tangent is ππ¦/ππ₯ Finding π π/π π π₯^2+π¦^2 β2π₯ β3=0 Differentiating w.r.t.π₯ π(π₯^2 + π¦^2 β2π₯ β3)/ππ₯=0 π(π₯^2 )/ππ₯+π(π¦^2 )/ππ₯βπ(2π₯)/ππ₯βπ(3)/ππ₯=0 2π₯+π(π¦^2 )/ππ¦ Γ ππ¦/ππ₯β2β0=0 π(π¦^2 )/ππ¦ Γ ππ¦/ππ₯=2β2π₯ 2π¦ Γ ππ¦/ππ₯=2β2π₯ ππ¦/ππ₯=(2 β 2π₯)/2π¦ ππ¦/ππ₯=(2 (1 β π₯))/2π¦ ππ¦/ππ₯=(1 β π₯)/π¦ Now, Since line is parallel to π₯βππ₯ππ Angle with π₯βππ₯ππ =0 π=0 Slope of π₯βππ₯ππ =tanβ‘π=tanβ‘0Β°=0 Now Slope of tangent = Slope of π₯βππ₯ππ ππ¦/ππ₯=0 (1 β π₯)/π¦=0 1βπ₯=0 Γ y 1βπ₯=0 π₯=1 Finding y when π₯=1 π₯^2+π¦^2β2π₯β3=0 (1)^2+π¦^2β2(1)β3=0 1+π¦^2β2β3=0 π¦^2β4=0 π¦^2=4 π¦=Β±β4 π¦=Β±2 Hence, the required points are (π , π) & (π , βπ)