If the tangent drawn to the curve (x^2+1)(y-3 | vedclass

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(B) Given the curve equation: $(x^2+1)(y-3)=x$ ... $(i)$ Differentiating both sides with respect to $x$: $(x^2+1) \frac{dy}{dx} + (2x)(y-3) = 1$ $\fra

Vedclass · Accepted Answer

$x=1$

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(B) Given the curve equation: $(x^2+1)(y-3)=x$ ... $(i)$ Differentiating both sides with respect to $x$: $(x^2+1) \frac{dy}{dx} + (2x)(y-3) = 1$ $\frac{dy}{dx} = \frac{1 - 2x(y-3)}{x^2+1}$ Since the tangent is a horizontal line,the slope $\frac{dy}{dx} = 0$. Therefore,$1 - 2x(y-3) = 0 \Rightarrow 2x(y-3) = 1$ ... $(ii)$ From $(i)$,we have $(y-3) = \frac{x}{x^2+1}$. Substituting this into $(ii)$: $2x \left( \frac{x}{x^2+1} \right) = 1$ $2x^2 = x^2 + 1 \Rightarrow x^2 = 1$. Since $P$ lies in the first quadrant,$x = 1$. Substituting $x=1$ into $(i)$: $(1^2+1)(y-3) = 1 \Rightarrow 2(y-3) = 1 \Rightarrow y-3 = \frac{1}{2} \Rightarrow y = \frac{7}{2}$. The point $P$ is $(1, \frac{7}{2})$. Since the tangent is horizontal,the normal is a vertical line passing through $x=1$. The equation of the normal is $x = 1$.

If the tangent drawn to the curve $(x^2+1)(y-3)=x$ at a point $P$,lying in the first quadrant,is a horizontal line,then the equation of the normal at the point $P$ is

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