Question:
The equation of the tangent to the conic $x^{2}-y^{2}-8x+2y+11=0$ at $(2,1)$ is
The equation of the tangent to the conic $x^{2}-y^{2}-8x+2y+11=0$ at $(2,1)$ is
Updated On: Apr 26, 2024
- $x + 2 = 0$
- $2 x +1 = 0$
- $x + y +1 = 0$
- $x - 2 = 0$
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The Correct Option is D
Solution and Explanation
Equation of the tangent at $\left(x_{1}, y_{1}\right)$ is
$xx_{1}-yy_{1}-4\left(x+x_{1}\right)+\left(y+y_{1}\right)+11=0$
Put $x_{1}=2$ and $ y_{1}=1$, we get
$2x - y - 4\left(x + 2\right)+ \left(y + 1\right)+ 11 = 0$
$\Rightarrow - 2x - 8 + 12= 0$
$\Rightarrow x-2=0$
$xx_{1}-yy_{1}-4\left(x+x_{1}\right)+\left(y+y_{1}\right)+11=0$
Put $x_{1}=2$ and $ y_{1}=1$, we get
$2x - y - 4\left(x + 2\right)+ \left(y + 1\right)+ 11 = 0$
$\Rightarrow - 2x - 8 + 12= 0$
$\Rightarrow x-2=0$
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