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

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

Vedclass · Accepted Answer

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

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(B) Given the hyperbola $x^2-y^2=9$ and the chord of contact $x=9$. Substituting $x=9$ into the hyperbola equation: $81-y^2=9$ $\Rightarrow y^2=72$ $\Rightarrow y = \pm 6\sqrt{2}$. Thus,the points of contact are $P_1(9, 6\sqrt{2})$ and $P_2(9, -6\sqrt{2})$. Differentiating $x^2-y^2=9$ with respect to $x$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}$. At $P_1(9, 6\sqrt{2})$,the slope $m_1 = \frac{9}{6\sqrt{2}} = \frac{3}{2\sqrt{2}}$. The equation of the tangent at $P_1$ is $y - 6\sqrt{2} = \frac{3}{2\sqrt{2}}(x-9)$,which simplifies to $2\sqrt{2}y - 24 = 3x - 27$,or $3x - 2\sqrt{2}y - 3 = 0$. At $P_2(9, -6\sqrt{2})$,the slope $m_2 = \frac{9}{-6\sqrt{2}} = -\frac{3}{2\sqrt{2}}$. The equation of the tangent at $P_2$ is $y + 6\sqrt{2} = -\frac{3}{2\sqrt{2}}(x-9)$,which simplifies to $2\sqrt{2}y + 24 = -3x + 27$,or $3x + 2\sqrt{2}y - 3 = 0$. 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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