Ex 6.3,24

Transcript

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﷯ 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﷯ −1﷯﷮𝑑𝑥﷯ 1﷮ 𝑏﷮2﷯﷯ 𝑑 𝑦﷮2﷯﷯﷮𝑑𝑥﷯= 𝑑﷮𝑑𝑥﷯ 𝑥﷮ 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﷯﷯﷮ 2𝑦﷮ 𝑏﷮2﷯﷯﷯ 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 = −1﷮ 𝑏﷮2﷯𝑥0﷮ 𝑎﷮2﷯𝑦0﷯﷯ Slope of Normal = − 𝑎﷮2﷯𝑦0﷮ 𝑏﷮2﷯𝑥0﷯ Finding equation of tangent & normal