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

Concept:

Equation of tangent at the point (x1, y1 ) of the curve y = f(x) is \(\rm (y-y_1) = m (x - x_1)\), where m = f'(x) is slope of the curve.

The line is passing through the point (x1, y1 ) if the point (x1, y1 ) is satisfying the equation of a line.

Calculator:

Given, equation of curve is x2y2 - 2x = 4(1 - y) 

⇒ \(\rm 2xy^2+2x^2y\dfrac {dy}{dx} - 2 = -4\dfrac {dy}{dx}\)

⇒ \(\rm \dfrac {dy}{dx} (2x^2y+4)=2 -2xy^2\)

⇒ \(\rm \dfrac {dy}{dx} =\dfrac {2 -2xy^2}{2x^2y+4}\)

⇒ \(\rm (\dfrac {dy}{dx})_{(2, - 2)} =\dfrac {2 -2(2)(-2)^2}{2(2)^2(-2)+4}\)

⇒ \(\rm (\dfrac {dy}{dx})_{(2, - 2)} =\dfrac 76\)

Equation of tangent at the point (x_1, y₁ ) of the curve y = f(x) is \(\rm (y-y_1) = m (x - x_1)\)where m is slope of the curve.

Here (x1, y1 ) =  (2, -2) and m = \(\rm \dfrac 7 6\)

⇒ \(\rm (y+2) = \dfrac 76 (x - 2)\) 

⇒ 7x - 6y = 26.

Hence, The equation of tangent at the point (2, -2) to the curve x2y2 - 2x = 4(1 - y) is  7x - 6y = 26.

Now, put the point (-2, -7) in The equation of tangent 7x - 6y = 26.

LHS = 7(-2) - 6(- 7) = 28

RHS = 26

⇒LHS \(\neq\) RHS

The tangent at the point (2, -2) to the curve x2y2 - 2x = 4(1 - y) does not pass through the point (-2, -7)