# Let the normal at a point P on the curve $y^{2}-3x^{2}+y+10=0$ intersect the y-axis at $\left ( 0,\frac{3}{2} \right ).$ If m is the slope of the tangent at P to the curve, then $\left | m \right |$ is equal to Option: 1 4 Option: 2 3 Option: 3 2 Option: 4 1

Equation of Straight Line (Part 1) -

Equation of Straight Line

(b) Point-Slope form

Let the equation of give line l with slope ‘m’ is

y = mx + c    …..(i)

(x1,y1) lies on the line i

y1= mx1+c   ……(ii)

From (i) and (ii) [(ii) - (i)]

y - y= m( x - x1)

The equation of a straight line whose slope is given as ‘m’ and passes through the point (x1,y1) is  .

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$\begin{array}{c}{2 y y^{\prime}+y^{\prime}-6 x=0} \\ {y^{\prime}=\frac{6 x}{2 y+1}} \\ {\frac{-1}{y^{\prime}}=\frac{-(2 y+1)}{6 x}}(\text{slope of the normal})\end{array}$

Equation of the normal $y-y_{1}=\frac{-\left(2 y_{1}+1\right)}{6 x_{1}}\left(x-x_{1}\right)$

Normal intersect at (0,3/2)

$\\\frac{3}{2}-y_{1}=\frac{-\left(2 y_{1}+1\right)}{6 x_{1}}\left(0-x_{1}\right)\\ 8y_1-8=0\\ y_1=1\\ x_1=\pm2\\ |m|=4$

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