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(D) Given equations are:$ax \sec 	heta + by 	an 	heta = a$  $(1)$$ax 	an 	heta + by \sec 	heta = b$  $(2)$Squaring both equations:$(ax \sec 	he

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$A$ hyperbola

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(D) Given equations are:$ax \sec 	heta + by 	an 	heta = a$  $(1)$$ax 	an 	heta + by \sec 	heta = b$  $(2)$Squaring both equations:$(ax \sec 	heta + by 	an 	heta)^2 = a^2$$(ax 	an 	heta + by \sec 	heta)^2 = b^2$Subtracting the second from the first:$(ax \sec 	heta + by 	an 	heta)^2 - (ax 	an 	heta + by \sec 	heta)^2 = a^2 - b^2$$a^2 x^2 (\sec^2 	heta - 	an^2 	heta) + b^2 y^2 (	an^2 	heta - \sec^2 	heta) = a^2 - b^2$Since $\sec^2 	heta - 	an^2 	heta = 1$,we have:$a^2 x^2 - b^2 y^2 = a^2 - b^2$This is the equation of a hyperbola.

The locus of the point of intersection of the lines $ax \sec \theta + by \tan \theta = a$ and $ax \tan \theta + by \sec \theta = b$,where $\theta$ is the parameter,is

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