A tangent to the curve 9 b^2 x^2 - 4 a^2 y^2 | vedclass

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Question

(C) The given equation of the hyperbola is $9 b^2 x^2 - 4 a^2 y^2 = 36 a^2 b^2$,which can be rewritten as $\frac{x^2}{4 a^2} - \frac{y^2}{9 b^2} = 1$.

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$4 x^2 - 9 y^2 = 1$

Vedclass · Answer

(C) The given equation of the hyperbola is $9 b^2 x^2 - 4 a^2 y^2 = 36 a^2 b^2$,which can be rewritten as $\frac{x^2}{4 a^2} - \frac{y^2}{9 b^2} = 1$. Let the point of tangency be $(x_0, y_0) = (2 a \sec 	heta, 3 b 	an 	heta)$. The equation of the tangent at $(x_0, y_0)$ is $\frac{x x_0}{4 a^2} - \frac{y y_0}{9 b^2} = 1$. Substituting the point,we get $\frac{x (2 a \sec 	heta)}{4 a^2} - \frac{y (3 b 	an 	heta)}{9 b^2} = 1$,which simplifies to $\frac{x \sec 	heta}{2 a} - \frac{y 	an 	heta}{3 b} = 1$. The intercepts on the axes are $x = 2 a \cos 	heta$ and $y = -3 b \cot 	heta$. Given that the intercepts are of unit length,we have $|2 a \cos 	heta| = 1$ and $|-3 b \cot 	heta| = 1$. Thus,$\cos 	heta = \frac{1}{2 a}$ and $\cot 	heta = -\frac{1}{3 b}$. Using the identity $\csc^2 	heta - \cot^2 	heta = 1$,we have $\sec^2 	heta = 1 + 	an^2 	heta$,or $\frac{1}{\cos^2 	heta} - \frac{1}{\cot^2 	heta} = 1$. Substituting the values,$(2 a)^2 - (-3 b)^2 = 1$,which gives $4 a^2 - 9 b^2 = 1$. Therefore,the locus of the point $(a, b)$ is $4 x^2 - 9 y^2 = 1$.

$A$ tangent to the curve $9 b^2 x^2 - 4 a^2 y^2 = 36 a^2 b^2$ makes intercepts of unit length on each of the coordinate axes. Then the point $(a, b)$ lies on:

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