Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

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Ex 6.3, 13 Find points on the curve π₯^2/9 + π¦^2/16 = 1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis. π₯^2/9 + π¦^2/16 = 1 π¦^2/16=1βπ₯^2/9 Differentiating w.r.t. π₯ π(π¦^2/16)/ππ₯=π(1β π₯^2/9)/ππ₯ 1/16 π(π¦^2 )/ππ₯=π(1)/ππ₯βπ(π₯^2/9)/ππ₯ 1/16 Γ π(π¦^2 )/ππ₯ Γ ππ¦/ππ¦=0β1/9 π(π₯^2 )/ππ₯ 1/16 Γ π(π¦^2 )/ππ¦ Γ ππ¦/ππ₯=(β 1)/9 π(π₯^2 )/ππ₯ 1/16 Γ2π¦ Γππ¦/ππ₯=(β 1)/( 9) 2π₯ ππ¦/ππ₯=((β 1)/( 9) 2π₯)/(1/16 2π¦) ππ¦/ππ₯=(β 16)/9 π₯/π¦ Hence ππ¦/ππ₯=(β 16)/9 π₯/π¦ parallel to π₯βππ₯ππ  Given tangent is parallel to π₯βππ₯ππ  β Slope of tangent = Slope of π₯βππ₯ππ  ππ¦/ππ₯=0 (β 16)/( 9) π₯/π¦=0 This is only possible if π₯=0 when π₯=0 π₯^2/4 + π¦^2/16=1 0/4+π¦^2/16=1 (πΌπ π¦=0, (β16)/( 9) π₯/0=β) π¦^2/16=1 π¦^2=16 π¦=Β±4 Hence the points are (π , π) & (π , βπ) parallel to π¦βππ₯ππ  Similarly if line is parallel to π¦βππ₯ππ  Angle with π₯βππ₯ππ  =90Β° ΞΈ = 90Β° Slope = tan ΞΈ = tan 90Β°=β Hence ππ¦/ππ₯=β 16/9 π₯/π¦=β 16π₯/9π¦=1/0 This will be possible only if Denominator is 0 9π¦=0 π¦=0 Now it is given that π₯^2/9+π¦^2/16=1 Putting π¦=0 π₯^2/9+0/16=1 π₯^2/9=1 π₯^2=9 π₯=β9 π₯=Β±3 Hence the points at which tangent is parallel to π¦βππ₯ππ  are (π , π) & (βπ , π)