KCET · Maths · Application of Derivatives
The equation of the normal to the curve \( y\left(1+x^{2}\right)=2-x \) where the tangent crosses \( x \)-axis is
- A \( 5 x-y-10=0 \)
- B \( x-5 y-10=0 \)
- C \( 5 x+y+10=0 \)
- D \( x+5 y+10=0 \)
Answer & Solution
Correct Answer
(A) \( 5 x-y-10=0 \)
Step-by-step Solution
Detailed explanation
Given curve, \(y\left(1+x^{2}\right)=2-x \rightarrow(1)\)
At x-axis, \(y=0\) so Eq. (1) becomes
\(\Rightarrow 0=2-x \Rightarrow x=2\)
So, the point is \((2,0)\)
Now, differentiate Eq. (1) with respect to x, we get
\(y^{\prime}\left(1+x^{2}\right)+y(2 x)=-1\)
At point \((2,0)\), we have
\(\Rightarrow y^{\prime}\left(1+2^{2}\right)+0(2 \times 2)=-1\)
\(\Rightarrow y^{\prime}(5)=-1 \Rightarrow y^{\prime}=\frac{-1}{5}\)
This is the slope of the tangent at point \((2,0)\)
So, slope of normal is
\(-\frac{1}{y^{\prime}}=5\)
Therefore, equation of normal is,
\(y-0=5(x-2)\)
\(\Rightarrow y=5 x-10\)
\(\Rightarrow 5 x-y-10=0\)
At x-axis, \(y=0\) so Eq. (1) becomes
\(\Rightarrow 0=2-x \Rightarrow x=2\)
So, the point is \((2,0)\)
Now, differentiate Eq. (1) with respect to x, we get
\(y^{\prime}\left(1+x^{2}\right)+y(2 x)=-1\)
At point \((2,0)\), we have
\(\Rightarrow y^{\prime}\left(1+2^{2}\right)+0(2 \times 2)=-1\)
\(\Rightarrow y^{\prime}(5)=-1 \Rightarrow y^{\prime}=\frac{-1}{5}\)
This is the slope of the tangent at point \((2,0)\)
So, slope of normal is
\(-\frac{1}{y^{\prime}}=5\)
Therefore, equation of normal is,
\(y-0=5(x-2)\)
\(\Rightarrow y=5 x-10\)
\(\Rightarrow 5 x-y-10=0\)
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