If the normal drawn at the point P on the cur | vedclass

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(A) Let the point $P$ be $(x_1, y_1)$. The curve is $y^2 = x^3 - x + 1$. Differentiating with respect to $x$, we get $2y \frac{dy}{dx} = 3x^2 - 1$, so

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$x - y = 0$

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(A) Let the point $P$ be $(x_1, y_1)$. The curve is $y^2 = x^3 - x + 1$. Differentiating with respect to $x$, we get $2y \frac{dy}{dx} = 3x^2 - 1$, so $\frac{dy}{dx} = \frac{3x_1^2 - 1}{2y_1}$.The slope of the normal at $P$ is $m_n = -\frac{1}{dy/dx} = -\frac{2y_1}{3x_1^2 - 1}$.Since the normal makes equal intercepts on the axes, its slope must be $\pm 1$. Given the curve, we test $m_n = -1$ (as the normal equation $x+y=c$ or $x-y=c$ implies slope $\pm 1$).If $m_n = -1$, then $\frac{2y_1}{3x_1^2 - 1} = 1 \implies 2y_1 = 3x_1^2 - 1$.Substituting $y_1^2 = x_1^3 - x_1 + 1$ and $y_1 = \frac{3x_1^2 - 1}{2}$, we get $(\frac{3x_1^2 - 1}{2})^2 = x_1^3 - x_1 + 1$. Solving this, we find $x_1 = 1$, which gives $y_1 = 1$. Thus $P = (1, 1)$.The slope of the tangent at $(1, 1)$ is $m_t = \frac{3(1)^2 - 1}{2(1)} = \frac{2}{2} = 1$.The equation of the tangent is $y - 1 = 1(x - 1)$, which simplifies to $y = x$, or $x - y = 0$.

If the normal drawn at the point $P$ on the curve $y^2 = x^3 - x + 1$ makes equal intercepts on the coordinate axes, then the equation of the tangent drawn to the curve at $P$ is

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