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Need explanation for: The normal to the curve y(x−2)(x−3)=x+6 at the point where the curve intersects the y-axis passes through the point :

The normal to the curve y(x−2)(x−3)=x+6 at the point where the curve intersects the y-axis passes through the point :

 

  • Option 1)

    \left ( \frac{1}{2} ,\frac{1}{2}\right )

  • Option 2)

    \left ( \frac{1}{2},-\frac{1}{3} \right )

  • Option 3)

    \left ( \frac{1}{2},\frac{1}{3} \right )

  • Option 4)

    \left ( -\frac{1}{2},-\frac{1}{2} \right )

 
Answers (1)
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As we learnt in 

Equation of Normal -

Equation of normal to the curve  y = f(x) at the point  P(x1, y1) on the curve having a slope  MN  is 

(y-y_{1})=M_{N}(x-x_{1})


=\frac{-1}{\frac{dy}{dx}_{(x_{1},y_{1})}}(x-x_{1})

-

 

y=\frac{x+6}{x^{2}-5x+6}\, \, \, \, \, \, \, \, \, \, \, \, point (0,1)

\frac{dy}{dx}=\frac{\left ( x^{2}-5x+6 \right )\times 1-\left ( x+6 \right )\left ( 2x-5 \right )}{\left ( x-2 \right )^{2}\left ( x-3 \right )^{2}}

=\frac{6+30}{36}=1=MT

M_{N}=-1

\therefore equation of normal 

y-1=-1\left ( x-0 \right )=-x+0

\therefore x+y=1 

Correct option is 1.

 


Option 1)

\left ( \frac{1}{2} ,\frac{1}{2}\right )

This option is correct.

Option 2)

\left ( \frac{1}{2},-\frac{1}{3} \right )

This option is incorrect.

Option 3)

\left ( \frac{1}{2},\frac{1}{3} \right )

This option is incorrect.

Option 4)

\left ( -\frac{1}{2},-\frac{1}{2} \right )

This option is incorrect.

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