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Ex 6.3, 24 - Find equations of tangent and normal to hyperbola

Ex 6.3,24 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,24 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,24 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.3,24 - Chapter 6 Class 12 Application of Derivatives - Part 5

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Ex 6.3, 24 Find the equations of the tangent and normal to the hyperbola 𝑥^2/𝑎^2 – 𝑦^2/𝑏^2 = 1 at the point (𝑥0 , 𝑦0)We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 Finding 𝒅𝒚/𝒅𝒙 𝑥^2/𝑎^2 −𝑦^2/𝑏^2 =1 −𝑦^2/𝑏^2 =1−𝑥^2/𝑎^2 𝑦^2/𝑏^2 =𝑥^2/𝑎^2 −1 Differentiating w.r.t.𝑥 𝑑(𝑦^2/𝑏^2 )/𝑑𝑥=𝑑(𝑥^2/𝑎^2 −1)/𝑑𝑥 1/𝑏^2 𝑑(𝑦^2 )/𝑑𝑥=𝑑/𝑑𝑥 (𝑥^2/𝑎^2 )−𝑑(1)/𝑑𝑥 1/𝑏^2 × 𝑑(𝑦^2 )/𝑑𝑥 × 𝑑𝑦/𝑑𝑦=1/𝑎^2 𝑑(𝑥^2 )/𝑑𝑥−0 1/𝑏^2 𝑑(𝑦^2 )/𝑑𝑦 × 𝑑𝑦/𝑑𝑥=1/𝑎^2 . 2𝑥 1/𝑏^2 ×2𝑦 × 𝑑𝑦/𝑑𝑥=1/𝑎^2 2𝑥 𝑑𝑦/𝑑𝑥=2𝑥/𝑎^2 × 𝑏^2/2𝑦 𝑑𝑦/𝑑𝑥=(𝑏^2 𝑥)/(𝑎^2 𝑦) Slope of tangent at (𝑥0 , 𝑦0) is 〖𝑑𝑦/𝑑𝑥│〗_((𝑥0 , 𝑦0) )=(𝑏^2 𝑥0)/(𝑎^2 𝑦0) We know that Slope of tangent × Slope of Normal =−1 (𝑏^2 𝑥0)/(𝑎^2 𝑦0) × Slope of Normal =−1 Slope of Normal = (−〖 𝑎〗^2 𝑦0)/(𝑏^2 𝑥0) 𝑑𝑦/𝑑𝑥=(𝑏^2 𝑥)/(𝑎^2 𝑦) Slope of tangent at (𝑥0 , 𝑦0) is 〖𝑑𝑦/𝑑𝑥│〗_((𝑥0 , 𝑦0) )=(𝑏^2 𝑥0)/(𝑎^2 𝑦0) We know that Slope of tangent × Slope of Normal =−1 (𝑏^2 𝑥0)/(𝑎^2 𝑦0) × Slope of Normal =−1 Slope of Normal = (−〖 𝑎〗^2 𝑦0)/(𝑏^2 𝑥0) Finding equation of tangent & normal We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent at (𝑥0 , 𝑦0) & having Slope (𝑏^2 𝑥0)/(𝑎^2 𝑦0) is (𝑦−𝑦0)=(𝑏^2 𝑥0)/(𝑎^2 𝑦0) (𝑥−𝑥0) 𝑎^2 𝑦0(𝑦−𝑦0)=𝑏^2 𝑥0 (𝑥− 𝑥0) 𝑎^2 (𝑦0𝑦−𝑦0^2 )=𝑏^2 (𝑥0 𝑥−𝑥0^2 ) (𝑦0 𝑦 − 𝑦0^2)/𝑏^2 =((𝑥0 𝑥 − 𝑥0^2 ))/𝑎^2 (𝑦0 𝑦 )/𝑏^2 −(𝑦0^2)/𝑏^2 =(𝑥0 𝑥 )/𝑎^2 −(𝑥0^2)/𝑎^2 Equation of Normal at (𝑥0 ,𝑦0) & having Slope(−〖 𝑎〗^2 𝑦0)/(𝑏^2 𝑥0) is (𝑦−𝑦0)=(−〖 𝑎〗^2 𝑦0)/(𝑏^(2 ) 𝑥0) (𝑥−𝑥0) ((𝑦 − 𝑦0))/(〖 𝑎〗^(2 ) 𝑦0)=(− 1)/(𝑏^(2 ) 𝑥0) (𝑥−𝑥0) (𝑦 − 𝑦0)/(〖 𝑎〗^(2 ) 𝑦0)=(− (𝑥 − 𝑥0))/(𝑏^(2 ) 𝑥0) (𝒚 − 𝒚𝟎)/(〖 𝒂〗^(𝟐 ) 𝒚𝟎)+(𝒙 −𝒙𝟎)/(〖 𝒃〗^(𝟐 ) 𝒙𝟎)=𝟎 (𝑦0 𝑦 )/𝑏^2 −(𝑥0 𝑥 )/𝑎^2 =−(𝑥0^2)/𝑎^2 +(𝑦0^2)/𝑏^2 ((𝑦0 𝑦 )/𝑏^2 −(𝑥0 𝑥 )/𝑎^2 )=−((𝑥0^2)/𝑎^2 −(𝑦0^2)/𝑏^2 ) ((𝑦0 𝑦 )/𝑏^2 −(𝑥0 𝑥 )/𝑎^2 )=−1 (𝒙𝟎 𝒙 )/𝒂^𝟐 −(𝒚𝟎 𝒚 )/𝒃^𝟐 =𝟏 Since point (𝑥_0 ,𝑦_0 ) lie on the Curve ∴ It will satisfy the Equation of Curve ∴ (𝑥0^2)/𝑎^2 −(𝑦0^2)/𝑏^2 =1

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.