![Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 2](https://d1avenlh0i1xmr.cloudfront.net/b0dc4979-4cc7-4007-a9ac-e72f51a3a942/slide9.jpg)
![Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 3](https://d1avenlh0i1xmr.cloudfront.net/5194f14f-4656-4811-99c7-0bd9d59e46ab/slide10.jpg)
![Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 4](https://d1avenlh0i1xmr.cloudfront.net/2697c7fa-e11b-4122-8aad-24ecde0952ec/slide11.jpg)
![Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 5](https://d1avenlh0i1xmr.cloudfront.net/ed1b2c89-fd18-40e7-aa8a-90778fb8bad3/slide12.jpg)
![Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 6](https://d1avenlh0i1xmr.cloudfront.net/c12a8ccc-df3d-42ad-b35d-af57f9105bb0/slide13.jpg)
![Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives - Part 7](https://d1avenlh0i1xmr.cloudfront.net/210403d1-7102-438a-a4aa-59e4734995d2/slide14.jpg)
Tangents and Normals (using Differentiation)
Tangents and Normals (using Differentiation)
Last updated at April 16, 2024 by Teachoo
Question 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=1βπ₯^2/9 Differentiating w.r.t. π₯ π(π¦^2/16)/ππ₯=π(1β π₯^2/9)/ππ₯ 1/16 π(π¦^2 )/ππ₯=π(1)/ππ₯βπ(π₯^2/9)/ππ₯ 1/16 Γ π(π¦^2 )/ππ₯ Γ ππ¦/ππ¦=0β1/9 π(π₯^2 )/ππ₯ 1/16 Γ π(π¦^2 )/ππ¦ Γ ππ¦/ππ₯=(β 1)/9 π(π₯^2 )/ππ₯ 1/16 Γ2π¦ Γππ¦/ππ₯=(β 1)/( 9) 2π₯ ππ¦/ππ₯=((β 1)/( 9) 2π₯)/(1/16 2π¦) ππ¦/ππ₯=(β 16)/9 π₯/π¦ Hence ππ¦/ππ₯=(β 16)/9 π₯/π¦ parallel to π₯βππ₯ππ Given tangent is parallel to π₯βππ₯ππ β Slope of tangent = Slope of π₯βππ₯ππ ππ¦/ππ₯=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 (πΌπ π¦=0, (β16)/( 9) π₯/0=β) π¦^2/16=1 π¦^2=16 π¦=Β±4 Hence the points are (π , π) & (π , βπ) parallel to π¦βππ₯ππ Similarly if line is parallel to π¦βππ₯ππ Angle with π₯βππ₯ππ =90Β° ΞΈ = 90Β° Slope = tan ΞΈ = tan 90Β°=β Hence ππ¦/ππ₯=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₯=β9 π₯=Β±3 Hence the points at which tangent is parallel to π¦βππ₯ππ are (π , π) & (βπ , π)