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Maths-

General

Easy

Question

The tangent to the curve $3 x y to the power of 2 end exponent minus 2 x to the power of 2 end exponent y equals 1$ at (1,1) meets the curve again at the point‐

$(\frac{16}{5}, \frac{1}{20})$
$(- \frac{16}{5}, - \frac{1}{20})$
$(\frac{1}{20}, \frac{16}{5})$
$(- \frac{1}{20}, \frac{16}{5})$

hint Hint:

Given equation $3 x y to the power of 2 end exponent minus 2 x to the power of 2 end exponent y equals 1$ at (1,1)

$T o space f i n d space t h e space s l o p e space w e space s h o u l d space f i n d space t h e space d e r i v a t i v e space o f space t h e space c u r v e 6 x y fraction numerator d y over denominator d x end fraction space plus 3 y squared minus 4 x y minus 2 x squared fraction numerator d y over denominator d x end fraction equals 0 fraction numerator d y over denominator d x end fraction equals fraction numerator 4 x y minus 3 y squared over denominator 6 x y minus 2 x squared end fraction W e space n e e d space s l o p e space a t space left parenthesis 1 comma 1 right parenthesis space s u b space i n space t h e space d i f f e r e n t i a t e d space e q u a t i o n m equals 1 fourth W e space c a n space w r i t e space t h e space tan g e n t space e q u a t i o n space a s left parenthesis y minus 1 right parenthesis equals 1 fourth left parenthesis x minus 1 right parenthesis x equals 4 y minus 3 I t space w i l l space t o u c h space t h e space c u r v e space a t space left parenthesis l comma m right parenthesis s o space w e space c a n space s u b s t i t u t e space a b o v e space o b t a i n e d space e q u a t i o n n space i n space t h e space c u r v e w e space g e t space l equals 1 comma 1 comma fraction numerator negative 16 over denominator 5 end fraction m equals 1 comma 1 comma fraction numerator negative 1 over denominator 20 end fraction H e n c e space t h e space p o i n t space i s space left parenthesis fraction numerator negative 16 over denominator 5 end fraction comma fraction numerator negative 1 over denominator 20 end fraction right parenthesis$

The correct answer is: $left parenthesis negative fraction numerator 16 over denominator 5 end fraction comma negative fraction numerator 1 over denominator 20 end fraction right parenthesis$

Firstly for the given curve we find the slope of the tangent at the given point and then obtain tangent equation and then substitute in curve to solve for the point
Given equation $3 x y to the power of 2 end exponent minus 2 x to the power of 2 end exponent y equals 1$ at (1,1)

$T o space f i n d space t h e space s l o p e space w e space s h o u l d space f i n d space t h e space d e r i v a t i v e space o f space t h e space c u r v e 6 x y fraction numerator d y over denominator d x end fraction space plus 3 y squared minus 4 x y minus 2 x squared fraction numerator d y over denominator d x end fraction equals 0 fraction numerator d y over denominator d x end fraction equals fraction numerator 4 x y minus 3 y squared over denominator 6 x y minus 2 x squared end fraction W e space n e e d space s l o p e space a t space left parenthesis 1 comma 1 right parenthesis space s u b space i n space t h e space d i f f e r e n t i a t e d space e q u a t i o n m equals 1 fourth W e space c a n space w r i t e space t h e space tan g e n t space e q u a t i o n space a s left parenthesis y minus 1 right parenthesis equals 1 fourth left parenthesis x minus 1 right parenthesis x equals 4 y minus 3 I t space w i l l space t o u c h space t h e space c u r v e space a t space left parenthesis l comma m right parenthesis s o space w e space c a n space s u b s t i t u t e space a b o v e space o b t a i n e d space e q u a t i o n n space i n space t h e space c u r v e w e space g e t space l equals 1 comma 1 comma fraction numerator negative 16 over denominator 5 end fraction m equals 1 comma 1 comma fraction numerator negative 1 over denominator 20 end fraction H e n c e space t h e space p o i n t space i s space left parenthesis fraction numerator negative 16 over denominator 5 end fraction comma fraction numerator negative 1 over denominator 20 end fraction right parenthesis$

The final answer is option B

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