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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

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prateek

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