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

Differentiate \(x 2y 2 - 2x = 4 - 4y\): \(2xy 2 + 2x 2yy' - 2 = -4y'\) At (2, −2): \(16 - 16y' - 2 = -4y' 12y' = 14 y' = 7/6\) Tangent: \(y + 2 = 7/6(x - 2) 7x - 6y = 26\) Check the points: (−4, −9): −28 + 54 = 26. On the line. (4, 1/3): 28 − 2 = 26. On the line. (8, 5): 56 − 30 = 26. On the line. (−2, −7): −14 + 42 = 28 ≠ 26. Not on the line. Answer: (A) (−2, −7)

The tangent at the point (2, -2) to the... - Solved

Question 36

Mathematics Differentiation Hard

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

(A) (-2, -7)

(B) (-4, -9)

(C) \(\left(4, \frac 1 3 \right)\)

(D) (8, 5)

✓ Correct Answer: Option A Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Differentiate \(x^2y^2 - 2x = 4 - 4y\):

\(2xy^2 + 2x^2yy' - 2 = -4y'\)

At (2, −2): \(16 - 16y' - 2 = -4y' \Rightarrow 12y' = 14 \Rightarrow y' = \frac{7}{6}\)

Tangent: \(y + 2 = \frac{7}{6}(x - 2) \Rightarrow 7x - 6y = 26\)

Check the points:

(−4, −9): −28 + 54 = 26. On the line.
(4, 1/3): 28 − 2 = 26. On the line.
(8, 5): 56 − 30 = 26. On the line.
(−2, −7): −14 + 42 = 28 ≠ 26. Not on the line.

Answer: (A) (−2, −7)

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Question 36

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