The required point is?

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The required point is?

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Answer 1 · 2018-03-15T13:05:02

Answer is (4).

Explanation:

Let us find the slope of tangent on curve $x^{2} y^{2} - 2 x = 4 (1 - y)$ $x^2y^2-2x=4(1-y)$ at point $(2, - 2)$ $(2,-2)$ , which is given by its derivative at the point. Derivative is given by

$2 x y^{2} + 2 x^{2} y \frac{d y}{d x} - 2 = - 4 \frac{d y}{d x}$ $2xy^2+2x^2y(dy)/(dx)-2=-4(dy)/(dx)$

i.e. $\frac{d y}{d x} = \frac{2 (1 - x y^{2})}{2 (x^{2} y + 2)} = \frac{2 (1 - 2 \cdot {(- 2)}^{2})}{2 (2^{2} \cdot (- 2) + 2)} = \frac{7}{6}$ $(dy)/(dx)=(2(1-xy^2))/(2(x^2y+2))=(2(1-2*(-2)^2))/(2(2^2*(-2)+2))=7/6$

and equation of tangent is $y + 2 = \frac{7}{6} (x - 2)$ $y+2=7/6(x-2)$ or $6 y - 7 x + 26 = 0$ $6y-7x+26=0$ or $y = \frac{7 x - 26}{6}$ $y=(7x-26)/6$

It can be easily seen that while $(4, \frac{1}{3}), (8, 5)$ $(4,1/3),(8,5)$ and $(- 4, - 9)$ $(-4,-9)$ lie on tangent, $(- 2, - 7)$ $(-2,-7)$ does not lie on tangent.

Hence answer is (4).

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The required point is?

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