Let the normal at a point P on the curve $$y2-3x2+y+10=0$$ intersect the y $$-axis$$ at $$0$$,$$32$$. If m is the slope of the tangent at P to the curve, then |m| is equal to

Let $$P=x1$$,$$y1$$. Then equation of tangent at P $$is2yy$$'$$-6x+y$$'$$=0$$⇒y'$$=6x11+2y1$$∴slope of normal at $$P=32-y1-x1=-1+2y16x1$$⇒$$9-6y1=1+2y1$$ ⇒$$y1=1$$∴$$x1=$$±$$2$$∴ Slope of tangent $$=m=$$±$$123$$ $$=$$±$$4$$∴$$m=4$$

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

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Solution

$\begin{array}{l} L e t P = (x_{1}, y_{1}) . T h e n e q u a t i o n o f \tan g e n t a t P i s \\ 2 y y' - 6 x + y' = 0 \Rightarrow y' = (\frac{6 x_{1}}{1 + 2 y_{1}}) \\ ∴ s l o p e o f n o r m a l a t P = (\frac{\frac{3}{2} - y_{1}}{- x_{1}}) = - (\frac{1 + 2 y_{1}}{6 x_{1}}) \\ \Rightarrow 9 - 6 y_{1} = 1 + 2 y_{1} \Rightarrow y_{1} = 1 \\ ∴ x_{1} = \pm 2 \\ ∴ Slope of tangent = m = (\frac{\pm 12}{3}) \\ = \pm 4 \\ ∴ |m| = 4 \end{array}$

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Tangents and normals

Remember concepts with our Masterclasses.

Let the normal at a point P on the curve y2-3x2+y+10=0 intersect the y -axis at 0,32. If m is the slope of the tangent at P to the curve, then |m| is equal to

Let P=x1,y1. Then equation of tangent at P is2yy'-6x+y'=0⇒y'=6x11+2y1∴slope of normal at P=32-y1-x1=-1+2y16x1⇒9-6y1=1+2y1 ⇒y1=1∴x1=±2∴ Slope of tangent =m=±123 =±4∴m=4

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