KCET · Maths · Determinants
The two curves \( x^{3}-3 x y^{2}+2=0 \) and \( 3 x^{2} y-y^{3}=2 \)
- A touch each other
- B cut at right angle
- C cut at angle \( \frac{\pi}{3} \)
- D cut at angle \( \frac{\pi}{4} \)
Answer & Solution
Correct Answer
(B) cut at right angle
Step-by-step Solution
Detailed explanation
Given Curves,
\(x^{3}-3 x y^{2}+2=0 \rightarrow(1)\)
\(3 x^{2} y-y^{3}=2 \rightarrow(2)\)
Differentiating Eqs. (1) and (2) with respect to x, we have
\(3 x^{2}-3\left(y^{2}+x(2 y y)\right)=0\)
\(\Rightarrow x^{2}=y^{2}+2 x y y\)
\(\Rightarrow y=\frac{x^{2}-y^{2}}{2 x y}=m_{1}\) (let) \(\rightarrow(3)\)
\(3\left(x^{2} y+2 x y\right)-3 y^{2} y=0\)
\(\Rightarrow x^{2} y+2 x y-y^{2} y=0\)
\(\Rightarrow y=-\frac{2 x y}{x^{2}-y^{2}}=m_{2}\) (let) \(\rightarrow\) (4)
From Eqs. (3) and (4), we have
\(m 1 \cdot m_{2}=-1\)
Hence, these curves cut at \(90^{\circ}\)
\(x^{3}-3 x y^{2}+2=0 \rightarrow(1)\)
\(3 x^{2} y-y^{3}=2 \rightarrow(2)\)
Differentiating Eqs. (1) and (2) with respect to x, we have
\(3 x^{2}-3\left(y^{2}+x(2 y y)\right)=0\)
\(\Rightarrow x^{2}=y^{2}+2 x y y\)
\(\Rightarrow y=\frac{x^{2}-y^{2}}{2 x y}=m_{1}\) (let) \(\rightarrow(3)\)
\(3\left(x^{2} y+2 x y\right)-3 y^{2} y=0\)
\(\Rightarrow x^{2} y+2 x y-y^{2} y=0\)
\(\Rightarrow y=-\frac{2 x y}{x^{2}-y^{2}}=m_{2}\) (let) \(\rightarrow\) (4)
From Eqs. (3) and (4), we have
\(m 1 \cdot m_{2}=-1\)
Hence, these curves cut at \(90^{\circ}\)
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