A tangent to the hyperbola frac{x^2}{4} - fra | vedclass

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(D) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec 	heta, b 	an 	heta)$ is $\frac{x \se

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$\frac{4}{x^2} - \frac{2}{y^2} = 1$

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(D) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec 	heta, b 	an 	heta)$ is $\frac{x \sec 	heta}{a} - \frac{y 	an 	heta}{b} = 1$.Setting $y=0$,we get $x = a \cos 	heta$,so $P = (a \cos 	heta, 0)$.Setting $x=0$,we get $y = -b \cot 	heta$,so $Q = (0, -b \cot 	heta)$.Since $OPRQ$ is a rectangle with $O(0,0)$,the coordinates of $R$ are $(h, k) = (a \cos 	heta, -b \cot 	heta)$.From this,$\cos 	heta = \frac{h}{a}$ and $\cot 	heta = -\frac{k}{b}$.Using the identity $\csc^2 	heta - \cot^2 	heta = 1$,we have $\frac{1}{\sin^2 	heta} - \cot^2 	heta = 1$,which is $\frac{1}{1 - \cos^2 	heta} - \cot^2 	heta = 1$.Alternatively,$\frac{1}{\cos^2 	heta} = 1 + 	an^2 	heta$ is not helpful here. Let's use $\csc^2 	heta = 1 + \cot^2 	heta$.$\frac{1}{\sin^2 	heta} = 1 + \frac{k^2}{b^2} = \frac{b^2 + k^2}{b^2}$.Since $\cos^2 	heta = \frac{h^2}{a^2}$,then $\sin^2 	heta = 1 - \frac{h^2}{a^2} = \frac{a^2 - h^2}{a^2}$.Substituting these into $\frac{1}{\sin^2 	heta} = 1 + \frac{k^2}{b^2}$ gives $\frac{a^2}{a^2 - h^2} = \frac{b^2 + k^2}{b^2}$.This simplifies to $a^2 b^2 = (a^2 - h^2)(b^2 + k^2) = a^2 b^2 + a^2 k^2 - b^2 h^2 - h^2 k^2$.$a^2 k^2 - b^2 h^2 - h^2 k^2 = 0 \Rightarrow \frac{a^2}{h^2} - \frac{b^2}{k^2} = 1$.Given $a^2 = 4$ and $b^2 = 2$,the locus of $R(x, y)$ is $\frac{4}{x^2} - \frac{2}{y^2} = 1$.

$A$ tangent to the hyperbola $\frac{x^2}{4} - \frac{y^2}{2} = 1$ meets the $x-$axis at $P$ and the $y-$axis at $Q$. Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on

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