A tangent to the ellipse frac{x^2}{25}+frac{y | vedclass

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(C) The equation of any tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ is given by $\frac{x \cos 	heta}{5} + \frac{y \sin 	heta}{4} = 1$.T

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$\frac{25}{x^2}+\frac{16}{y^2}=4$

Vedclass · Answer

(C) The equation of any tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ is given by $\frac{x \cos 	heta}{5} + \frac{y \sin 	heta}{4} = 1$.The points of intersection with the coordinate axes are $A = (\frac{5}{\cos 	heta}, 0)$ and $B = (0, \frac{4}{\sin 	heta})$.Since $	riangle AOB$ is a right-angled triangle with the right angle at the origin $O(0,0)$,the circumcentre $(h, k)$ is the midpoint of the hypotenuse $AB$.Thus,$h = \frac{5}{2 \cos 	heta}$ and $k = \frac{4}{2 \sin 	heta}$.This implies $\cos 	heta = \frac{5}{2h}$ and $\sin 	heta = \frac{4}{2k} = \frac{2}{k}$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we get $(\frac{5}{2h})^2 + (\frac{2}{k})^2 = 1$,which simplifies to $\frac{25}{4h^2} + \frac{4}{k^2} = 1$.Multiplying by $4$,we get $\frac{25}{h^2} + \frac{16}{k^2} = 4$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{25}{x^2} + \frac{16}{y^2} = 4$.

$A$ tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ intersects the coordinate axes at $A$ and $B$. Then the locus of the circumcentre of triangle $AOB$ (where $O$ is the origin) is:

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