The equation of the tangent parallel to y-x+5 | vedclass

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Question

(A) The given hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$. Here,$a^{2}=3$ and $b^{2}=2$. The equation of the tangent parallel to $y-x+5=0$ is of

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$x-y-1=0$

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(A) The given hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$. Here,$a^{2}=3$ and $b^{2}=2$. The equation of the tangent parallel to $y-x+5=0$ is of the form $y=x+c$. Comparing with $y=mx+c$,we have $m=1$. The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}-b^{2}$. Substituting the values,$c^{2}=3(1)^{2}-2 = 3-2 = 1$. Thus,$c=\pm 1$. The equations of the tangents are $y=x+1$ or $y=x-1$,which can be written as $x-y+1=0$ or $x-y-1=0$. Comparing with the given options,$x-y-1=0$ is the correct choice.

The equation of the tangent parallel to $y-x+5=0$ drawn to $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$ is

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