CBSE Class 12-science Answered

Find the equations of the tangent and the normal to 16x² + 9y² = 144 at (x₁, y₁) where x₁ = 2 and y₁ > 0.

Asked by Topperlearning User | 07 Aug, 2014, 01:00: PM

Expert Answer

The equation of the given curve is

16x² + 9y² = 144 …(i)

Since (x₁, y₁) lies on (i). Therefore,

$16 straight x subscript 1 squared plus 9 straight y subscript 1 squared equals 144 rightwards double arrow 16 left parenthesis 2 right parenthesis squared plus 9 straight y subscript 1 squared equals 144 rightwards double arrow straight y subscript 1 squared equals 80 over 9 rightwards double arrow straight y subscript 1 equals fraction numerator 4 square root of 5 over denominator 3 end fraction left square bracket because straight y subscript 1 greater than 0 right square bracket$

So, coordinates of the given point are

open parentheses 2 comma fraction numerator 4 square root of 5 over denominator 3 end fraction close parentheses

Now,

$16 straight x squared plus 9 straight y squared equals 144 rightwards double arrow 32 straight x plus 18 straight y dy over dx equals 0 space left square bracket Differentiating space straight w. straight r. straight t. space straight x right square bracket rightwards double arrow dy over dx equals minus fraction numerator 16 straight x over denominator 9 straight y end fraction rightwards double arrow open parentheses dy over dx close parentheses subscript open parentheses 2 comma fraction numerator 4 square root of 5 over denominator 3 end fraction close parentheses end subscript equals minus fraction numerator 16 cross times 2 over denominator 9 cross times begin display style fraction numerator 4 square root of 5 over denominator 3 end fraction end style end fraction equals minus fraction numerator 8 over denominator 3 square root of 5 end fraction$

The equation of the tangent at

open parentheses 2 comma fraction numerator 4 square root of 5 over denominator 3 end fraction close parentheses

$straight y minus fraction numerator 4 square root of 5 over denominator 3 end fraction equals open parentheses dy over dx close parentheses subscript open parentheses 2 comma fraction numerator 4 square root of 5 over denominator 3 end fraction close parentheses end subscript left parenthesis straight x minus 2 right parenthesis rightwards double arrow straight y minus fraction numerator 4 square root of 5 over denominator 3 end fraction equals minus fraction numerator 8 over denominator 3 square root of 5 end fraction left parenthesis straight x minus 2 right parenthesis rightwards double arrow 8 straight x plus 3 square root of 5 minus 36 equals 0$

The equation of the normal at

open parentheses 2 comma fraction numerator 4 square root of 5 over denominator 3 end fraction close parentheses

$straight y minus fraction numerator 4 square root of 5 over denominator 3 end fraction equals minus 1 over open parentheses dy over dx close parentheses subscript open parentheses 2 comma fraction numerator 4 square root of 5 over denominator 3 end fraction close parentheses end subscript left parenthesis straight x minus 2 right parenthesis rightwards double arrow straight y minus fraction numerator 4 square root of 5 over denominator 3 end fraction equals minus fraction numerator 1 over denominator minus fraction numerator 8 over denominator 3 square root of 5 end fraction end fraction left parenthesis straight x minus 2 right parenthesis rightwards double arrow straight y minus fraction numerator 4 square root of 5 over denominator 3 end fraction equals fraction numerator 3 square root of 5 over denominator 8 end fraction left parenthesis straight x minus 2 right parenthesis rightwards double arrow 9 square root of 5 straight x minus 24 space straight y plus 14 square root of 5 equals 0$