If x=9 is a chord of contact of the hyperbola | vedclass

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(B) Given that $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$. Substituting $x=9$ into the hyperbola equation: $81-y^2=9$ $y^2=72$ $y = \pm

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$3 x+2 \sqrt{2} y-3=0$

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(B) Given that $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$. Substituting $x=9$ into the hyperbola equation: $81-y^2=9$ $y^2=72$ $y = \pm 6\sqrt{2}$. Thus,the points of contact are $(9, 6\sqrt{2})$ and $(9, -6\sqrt{2})$. Differentiating $x^2-y^2=9$ with respect to $x$: $2x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{y}$. At point $(9, 6\sqrt{2})$,the slope $m = \frac{9}{6\sqrt{2}} = \frac{3}{2\sqrt{2}}$. The equation of the tangent at $(9, 6\sqrt{2})$ is: $y - 6\sqrt{2} = \frac{3}{2\sqrt{2}}(x-9)$ $2\sqrt{2}y - 24 = 3x - 27$ $3x - 2\sqrt{2}y - 3 = 0$. At point $(9, -6\sqrt{2})$,the slope $m = \frac{9}{-6\sqrt{2}} = -\frac{3}{2\sqrt{2}}$. The equation of the tangent at $(9, -6\sqrt{2})$ is: $y + 6\sqrt{2} = -\frac{3}{2\sqrt{2}}(x-9)$ $2\sqrt{2}y + 24 = -3x + 27$ $3x + 2\sqrt{2}y - 3 = 0$. Comparing with the options,$3x + 2\sqrt{2}y - 3 = 0$ is option $B$.

If $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$,then the equation of the tangent at one of the points of contact is

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