The equation of the tangent to the conic x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is
(A) x + 2 = 0 (B) 2x + 1 = 0 (C) x + y + 1 = 0 (D) x – 2 = 0

Question

The equation of the tangent to the conic x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is (A) x + 2 = 0 (B) 2x + 1 = 0 (C) x + y + 1 = 0 (D) x – 2 = 0

Saurabh Koranglekar · Answer

Vikas TU · Answer

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The equation of the tangent to the conic x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is (A) x + 2 = 0 (B) 2x + 1 = 0 (C) x + y + 1 = 0 (D) x – 2 = 0

The equation of the tangent to the conic x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is
(A) x + 2 = 0 (B) 2x + 1 = 0 (C) x + y + 1 = 0 (D) x – 2 = 0

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The equation of the tangent to the conic x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is (A) x + 2 = 0 (B) 2x + 1 = 0 (C) x + y + 1 = 0 (D) x – 2 = 0

The equation of the tangent to the conic x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is (A) x + 2 = 0 (B) 2x + 1 = 0 (C) x + y + 1 = 0 (D) x – 2 = 0

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The equation of the tangent to the conic x2 – y2 – 8x + 2y + 11 = 0 at (2, 1) is
(A) x + 2 = 0 (B) 2x + 1 = 0 (C) x + y + 1 = 0 (D) x – 2 = 0