The equation of normal of x2
+ y2
– 2x + 4y – 5 = 0 at (2, 1) is
(A) y = 3x – 5 (B) 2y = 3x – 4 (C) y = 3x + 4 (D) y = x + 1

Question

The equation of normal of x2 + y2 – 2x + 4y – 5 = 0 at (2, 1) is (A) y = 3x – 5 (B) 2y = 3x – 4 (C) y = 3x + 4 (D) y = x + 1

Vikas TU · Answer

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The equation of normal of x2 + y2 – 2x + 4y – 5 = 0 at (2, 1) is (A) y = 3x – 5 (B) 2y = 3x – 4 (C) y = 3x + 4 (D) y = x + 1

The equation of normal of x2
+ y2
– 2x + 4y – 5 = 0 at (2, 1) is
(A) y = 3x – 5 (B) 2y = 3x – 4 (C) y = 3x + 4 (D) y = x + 1

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The equation of normal of x2 + y2 – 2x + 4y – 5 = 0 at (2, 1) is (A) y = 3x – 5 (B) 2y = 3x – 4 (C) y = 3x + 4 (D) y = x + 1

The equation of normal of x2 + y2 – 2x + 4y – 5 = 0 at (2, 1) is (A) y = 3x – 5 (B) 2y = 3x – 4 (C) y = 3x + 4 (D) y = x + 1

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The equation of normal of x2
+ y2
– 2x + 4y – 5 = 0 at (2, 1) is
(A) y = 3x – 5 (B) 2y = 3x – 4 (C) y = 3x + 4 (D) y = x + 1