X cos θ – y sin θ = a cos 4 θ – a sin 4 θ y sin θ – a sin 4 θ = x cos θ – a cos 4 θ Show that the equation of the normal to the curve x = a cos 3 θ, y = a sin 3 θ at ‘θ’ is x cos θ – y sin θ = a cos 2θ Similarly it can be proved at the other points also. Prove that the curves 2x 2 + 4y 2 = 1 and 6x 2 – 12y 2 = 1 cut each other at right angles. ![]() It is given that the normal is parallel to the line ⇒ m 1 = m 2 ![]() (ii) x = a cos 3 t, sin 3 t at t = \(\frac\) is equal to a.įind the equations of a normal to y = x 3 – 3x that is parallel to 2x + 18y – 9 = 0. Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2įind the slope of the tangent to the curves at the respective given points. ![]() You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.
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