Question

The tangent at the point (2, -2) to the curve, ${x^2} {y^2} - 2x = 4(1 - y) $ does not pass through the point.
A (8,5)
B $\left ({4, \dfrac {1}{3}} \right) $
C (-2, -7)
D (-4, -9)

Answer 1

Hint: In this question we have been given a tangent at the point (2, -2) to the curve, ${x^2}{y^2} - 2x = 4(1 - y)$, we need to find the point which does not pass through it so for that firstly we will be finding $\dfrac{{dy}}{{dx}}$ which is the slope, then we will keep the points (2,-2) in place of (x,y) and find the equation of the tangent by using the formula: $y - {y_1} = m(x - {x_1})$. After finding the equation we will keep all the given points one by one to check which one satisfies the equation.

Complete step by step answer:
We have been provided a curve ${x^2} {y^2} - 2x = 4(1 - y) $, firstly we will be finding the slope for forming the equation of the tangent,
So, slope of tangent= $\dfrac{{dy}} {{dx}} $,
Now we will be differentiating ${x^2} {y^2} - 2x = 4(1 - y) $ with respect to dx using the formula: $\dfrac{d}{{dx}} ({x^n}) = n {x^ {n - 1}} $,
So, the differentiation would be: $2x{y^2} + 2y\dfrac{{dy}} {{dx}} {x^2} - 2 = - 4\dfrac{{dy}} {{dx}} $,
Now take the term $\dfrac{{dy}} {{dx}} $ at one side: $\dfrac{{dy}} {{dx}} \left ({4 + 2y{x^2}} \right) = 2 - 2x{y^2} $,
Solving it further we get: $\dfrac{{dy}} {{dx}} = \dfrac {{2 - 2x{y^2}}} {{4 + 2y{x^2}}} $,
Now putting the values (2, -2) in place of (x, y) we have: $\dfrac{{dy}} {{dx}} = \dfrac {{2 - 2(2)(4)}} {{4 + 2(- 2) (4)}} $,
Solving it further we get: $\dfrac{{dy}} {{dx}} = \dfrac {{- 14}} {{- 12}} = \dfrac {7}{6} $,
Now we will be finding the equation of tangent using the formula: $y - {y_1} = m (x - {x_1}) $,
Where $({x_1}, {y_1}) = (2, - 2) $ and $m = \dfrac{{dy}} {{dx}} = \dfrac {7}{6} $,
Now putting these values, we will get: $y + 2 = \dfrac {7}{6} (x - 2) $,
Solving it further we get: $6(y + 2) = 7(x - 2) $,
So, the equation comes out to be: $7x - 6y = 26$,
Now we have been given 4 points in the questions, so we need to keep these points one by one to check whether it satisfies the equation or not.
Lt us start with (8,5): $7(8) - 6(5) = 56 - 30 = 26$,
Therefore we have: $26 = 26$,
Hence points (8,5) satisfies the equation $7x - 6y = 26$.
Now, let us consider $\left ({4, \dfrac {1}{3}} \right) $: $7(4) - 6\left ({\dfrac {1}{3}} \right) = 28 - 2 = 26$,
Therefore we have: $26 = 26$,
Hence points $\left ({4, \dfrac {1}{3}} \right) $ satisfies the equation $7x - 6y = 26$.
Now, let us check for $(- 2, - 7) $: $7(- 2) - 6(- 7) = - 14 + 42 = 28$,
Therefore we have: $28 \ne 26$,
Hence points $(- 2, - 7) $ does not satisfy the equation $7x - 6y = 26$.
Now, we will check for last option $(- 4, - 9) $: $7(- 4) - 6(- 9) = - 28 + 54 = 26$,
Therefore we have: $26 = 26$,
Hence points $(- 4, - 9) $ does not satisfy the equation $7x - 6y = 26$.

So, the correct answer is “Option C”.

Note: In this question, many students try to put the sub options but it is not the correct way to approach this question. Firstly, we need to find the tangent and then put the sub options. Also, while putting all the options for checking which satisfies the equation, be careful as chances of mistakes are more over there. Also, while differentiating use this formula: $\dfrac{d}{{dx}} ({x^n}) = n {x^ {n - 1}} $ just for n > 0 otherwise use $\dfrac{d}{{dx}} ({x^ {- 1}}) = \log (x)$.