The equation of the tangent to the conic x^2 | vedclass

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Question

(D) The equation of the tangent to the conic $S: x^2 - y^2 - 8x + 2y + 11 = 0$ at point $(x_1, y_1)$ is given by $T = 0$.Replacing $x^2 	o xx_1$,$y^2

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$x - 2 = 0$

Vedclass · Answer

(D) The equation of the tangent to the conic $S: x^2 - y^2 - 8x + 2y + 11 = 0$ at point $(x_1, y_1)$ is given by $T = 0$.Replacing $x^2 	o xx_1$,$y^2 	o yy_1$,$x 	o \frac{x+x_1}{2}$,and $y 	o \frac{y+y_1}{2}$:$xx_1 - yy_1 - 8\left(\frac{x+x_1}{2}ight) + 2\left(\frac{y+y_1}{2}ight) + 11 = 0$Substituting $(x_1, y_1) = (2, 1)$:$x(2) - y(1) - 4(x + 2) + 1(y + 1) + 11 = 0$$2x - y - 4x - 8 + y + 1 + 11 = 0$$-2x + 4 = 0$$2x = 4$$x = 2$Therefore,the equation is $x - 2 = 0$.

The equation of the tangent to the conic $x^2 - y^2 - 8x + 2y + 11 = 0$ at the point $(2, 1)$ is:

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