Let the normal at a point P on the curve y^{2 | vedclass

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(B) Let the point $P$ be $(\alpha, \beta)$. Since $P$ lies on the curve,we have $\beta^{2}-3\alpha^{2}+\beta+10=0 \dots(i)$.Differentiating the equati

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$4$

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(B) Let the point $P$ be $(\alpha, \beta)$. Since $P$ lies on the curve,we have $\beta^{2}-3\alpha^{2}+\beta+10=0 \dots(i)$.Differentiating the equation of the curve with respect to $x$,we get $2yy' - 6x + y' = 0$,which implies $y'(2y+1) = 6x$,so $y' = \frac{6x}{2y+1}$.The slope of the tangent $m$ at $P(\alpha, \beta)$ is $m = \frac{6\alpha}{2\beta+1} \dots(ii)$.The slope of the normal at $P$ is $-\frac{1}{m} = -\frac{2\beta+1}{6\alpha}$.The normal passes through $(0, \frac{3}{2})$ and $(\alpha, \beta)$,so its slope is $\frac{\beta - 3/2}{\alpha - 0} = \frac{2\beta-3}{2\alpha}$.Equating the slopes: $\frac{2\beta-3}{2\alpha} = -\frac{2\beta+1}{6\alpha}$.Assuming $\alpha 
eq 0$,we have $3(2\beta-3) = -(2\beta+1) \Rightarrow 6\beta - 9 = -2\beta - 1 \Rightarrow 8\beta = 8 \Rightarrow \beta = 1$.Substituting $\beta = 1$ into $(i)$: $1^{2} - 3\alpha^{2} + 1 + 10 = 0 \Rightarrow 3\alpha^{2} = 12 \Rightarrow \alpha^{2} = 4$.From $(ii)$,$|m| = |\frac{6\alpha}{2(1)+1}| = |\frac{6\alpha}{3}| = |2\alpha|$.Since $\alpha^{2} = 4$,$|\alpha| = 2$,thus $|m| = 2 	imes 2 = 4$.

Let the normal at a point $P$ on the curve $y^{2}-3x^{2}+y+10=0$ intersect the $y$-axis at $(0, \frac{3}{2})$. If $m$ is the slope of the tangent at $P$ to the curve,then $|m|$ is equal to

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