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If the tangent to the conic y-6=x^{2}  at (2, 10) touches the circle, x^{2}+y^{2}+8x-2y=k (for some fixed k) at a  point\alpha ,\beta;then\alpha ,\beta is:

  • Option 1)

    \left ( -\frac{6}{17} ,\frac{10}{17}\right )

  • Option 2)

    \left ( -\frac{8}{17} ,\frac{2}{17}\right )

  • Option 3)

    \left ( -\frac{4}{17} ,\frac{1}{17}\right )

  • Option 4)

    \left ( -\frac{7}{17} ,\frac{6}{17}\right )

 

Answers (1)

best_answer

As we learnt in

Equation of tangent -

xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0
 

- wherein

Tangent to circle

x^{2}+y^{2}+2gx+2fy+c=0  at  (x_{1},y_{1})

 

 y-6-x^{2}=0

tongent at (2, 10)

\left ( \frac{y+10}{2} \right )-6-2x=0

y-4x-2=0

4x-y+2=0 - - -- - - - - -\left ( 1 \right )

Tangent to x^{2}+y^{2}+8x-2y-k=0 at \left ( x_{1}y_{1} \right ) is

xx_{1}+yy_{1}+4\left ( x+x_{1} \right )-\left ( y+y_{1} \right )-k=0

\left ( x_{1}+4 \right )x+\left ( \left y_{1}-1 \right \right )y+\left ( 4x_{1}-y_{1} -k\right )=0 - -- - - -- - - \left ( 2 \right )

4x-y+2=0

Comparing these two equations, since they represent same tangent.

\frac{x_{1}+4}{4}=\frac{1-y_{1}}{1}=\frac{4x_{1}-y_{1}-k}{2}

clearly \left ( -\frac{8}{17},\frac{2}{17} \right ) satisfies this relation.


Option 1)

\left ( -\frac{6}{17} ,\frac{10}{17}\right )

Incorrect option

Option 2)

\left ( -\frac{8}{17} ,\frac{2}{17}\right )

Correct option

Option 3)

\left ( -\frac{4}{17} ,\frac{1}{17}\right )

Incorrect option

Option 4)

\left ( -\frac{7}{17} ,\frac{6}{17}\right )

Incorrect option

Posted by

prateek

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