The locus of the middle point of the intercep | vedclass

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Question

(c) Let the point of contact be

$R \equiv (\sqrt 2 \cos 	heta ,\,\sin 	heta )$

Equation of tangent $AB$ is

$\frac{x}{{\sqrt 2 }}\cos 	heta

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) Let the point of contact be

$R \equiv (\sqrt 2 \cos 	heta ,\,\sin 	heta )$

Equation of tangent $AB$ is

$\frac{x}{{\sqrt 2 }}\cos 	heta + y\sin 	heta = 1$

$ \Rightarrow $ $A \equiv (\sqrt 2 \sec 	heta ,\,0);\,B \equiv (0,\,{m{cosec }}	heta )$

Let the middle point $Q$ of $AB$ be $(h,\,k)$

$ \Rightarrow $ $h = \frac{{\sec 	heta }}{{\sqrt 2 }},\,k = \frac{{{m{cosec }}	heta }}{2} $

$\Rightarrow \cos 	heta = \frac{1}{{h\sqrt 2 }},\,\sin 	heta = \frac{1}{{2k}}$

$ \Rightarrow $ $\frac{1}{{2{h^2}}} + \frac{1}{{4{k^2}}} = 1$,

Required locus is $\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1$.

Trick : The locus of mid-points of the portion of tangents to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ intercepted between axes is

${a^2}{y^2} + {b^2}{x^2} = 4{x^2}{y^2}$

$i.e.$, $\frac{{{a^2}}}{{4{x^2}}} + \frac{{{b^2}}}{{4{y^2}}} = 1$

or $\frac{1}{{2{x^2}}} + \frac{1}{{4{y^2}}} = 1$.

The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse ${x^2} + 2{y^2} = 2$ between the co-ordinates axes, is

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