The tangent at the point (2, -2) to the curve, x2y2 - 2x = 4(1 - y) does not pass through …

Question

The tangent at the point (2, -2) to the curve, x2y2 - 2x = 4(1 - y) does not pass through the point :

Accepted Answer

As,    $${{dy} \over {dx}}$$ = - $$\left[ {{{{{\delta f} \over {\delta x}}} \over {{{\delta f} \over {\delta y}}}}} \right]$$

$${{{\delta f} \over {\delta x}}}$$ = y2 $$ \times $$2x - 2

$${{{\delta f} \over {\delta y}}}$$ = x2 $$ \times $$ 2y + 4

$$\therefore\,\,\,$$ $${{dy} \over {dx}}$$ = - $$\left( {{{2x{y^2} - 2} \over {2{x^2}y + 4}}} \right)$$

$${\left[ {{{dy} \over {dx}}} \right]_{(2, - 2)}}$$ = - $$\left( {{{2 \times 2 \times 4 - 2} \over {2 \times 4 \times ( - 2) + 4}}} \right)$$ = - $$\left( {{{14} \over { - 12}}} \right)$$ = $${7 \over 6}$$

$$\therefore\,\,\,$$ Slope of tangent to the curve = $${7 \over 6}$$

Equation of tangent passes through (2, - 2) is

y + 2 = $${7 \over 6}$$ (x - 2)

$$ \Rightarrow $$ $\,\,\,$7x$-$$ 6y = 26 . . . . .(1)

Now put each option in equation (1) and see which one does not satisfy the equation.

By verifying each points you can see ($$-$2,$-$

Solutions · 1