WBJEE · Maths · Differentiation
For the curve \(x^{2}+4 x y+8 y^{2}=64\) the tangents are parallel to the \(x\) -axis only at the points
- A \((0,2 \sqrt{2})\) and \((0,-2 \sqrt{2})\)
- B \((8, -4)\) and (-8,4)
- C \((8 \sqrt{2},-2 \sqrt{2})\) and \((-8 \sqrt{2}, 2 \sqrt{2})\)
- D (9,0) and (-8,0)
Answer & Solution
Correct Answer
(B) \((8, -4)\) and (-8,4)
Step-by-step Solution
Detailed explanation
Given curve is, \(x^{2}+4 x y+8 y^{2}=64\)...(i) On differentiating w.r.t \(x,\) we get \(\quad 2 x+4\left(y+x \frac{d y}{d x}\right)+16 y \frac{d y}{d x}=0\) \(\Rightarrow \quad 2 x+4 y+(4 x+16 y) \frac{d y}{d x}=0\)…
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