Application of Derivatives

Find the point on curve 4x 2 + 9y 2 = 1, where the tangents are perpendicular to the line 2y + x = 0. - Wired Faculty

Question

Find the point on curve 4x² + 9y² = 1, where the tangents are perpendicular to the line 2y + x = 0.

Solution

The equation of curve is 4x² + 9y² = 1 ...(1)
Diff. w.r.t. x, $8 straight x plus 18 straight y dy over dx space equals space 0$
$therefore space space space dy over dx space equals space minus 4 over 9. space straight x over straight y space space space space space space rightwards double arrow space space space space space space space space space space slope space of space tangent space space equals space minus fraction numerator 4 straight x over denominator 9 straight y end fraction therefore space space space space space space straight m subscript 1 space equals space minus 4 over 9 straight x over straight y comma space space space where space straight m subscript 1 space space is space slope space of space tangent. space Let space straight m subscript 2 space be space slope space of space space 2 straight y plus straight x space equals space 0 therefore space space space space space space straight m subscript 2 space equals space minus 1 half space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis because space space tangent space is space perpendicular space to space line space left parenthesis 2 right parenthesis therefore space space space space space space straight m subscript 1 straight m subscript 2 space equals space minus 1. space space space space space space space space space space space space space space space space space space rightwards double arrow space space fraction numerator 4 straight x over denominator 9 straight y end fraction cross times negative 1 half space equals space minus 1 space space space space space rightwards double arrow space space space space space straight y space equals space minus fraction numerator 2 straight x over denominator 9 end fraction Putting space this space value space of space straight y space in space left parenthesis 1 right parenthesis comma space we space get comma 4 straight x squared plus fraction numerator 4 straight x squared over denominator 9 end fraction equals 1 comma space space space space space space space space space space space space space space space space rightwards double arrow space space straight x squared space equals space 9 over 40 comma space space space space space space space space rightwards double arrow space space space space space straight x space space equals plus-or-minus fraction numerator 3 over denominator square root of 40 end fraction space equals space plus-or-minus fraction numerator 3 over denominator 2 square root of 10 end fraction therefore space space space space from space left parenthesis 3 right parenthesis comma space space space straight y space equals space minus 2 over 9 cross times fraction numerator 3 over denominator 2 square root of 10 end fraction comma space space fraction numerator negative 2 over denominator 9 end fraction cross times fraction numerator negative 3 over denominator 2 square root of 10 end fraction equals negative fraction numerator 1 over denominator 3 square root of 10 end fraction comma space fraction numerator 1 over denominator 3 square root of 10 end fraction therefore space space space space space points space are space open parentheses fraction numerator 3 over denominator 2 square root of 10 end fraction comma space space space space minus fraction numerator 1 over denominator 3 square root of 10 end fraction close parentheses space and space space open parentheses negative fraction numerator 3 over denominator 2 square root of 10 end fraction comma space space fraction numerator 1 over denominator 3 square root of 10 end fraction close parentheses$

For Daily Free Study Material Join wiredfaculty