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Maths-

General

Easy

Question

The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse $x to the power of 2 end exponent plus 2 y to the power of 2 end exponent equals 2$ between the co-ordinates axes, is

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

The correct answer is: $fraction numerator 1 over denominator 2 x to the power of 2 end exponent end fraction plus fraction numerator 1 over denominator 4 y to the power of 2 end exponent end fraction equals 1$

Let the point of contact be
$R identical to left parenthesis square root of 2 cos invisible function application theta comma sin invisible function application theta right parenthesis$
Equation of tangent AB is

$fraction numerator x over denominator square root of 2 end fraction cos invisible function application theta plus y sin invisible function application theta equals 1$
$rightwards double arrow A identical to left parenthesis square root of 2 sec invisible function application theta comma 0 right parenthesis semicolon B identical to left parenthesis 0 comma text cosec end text theta right parenthesis$
Let the middle point Q of AB be $left parenthesis h comma k right parenthesis$
$rightwards double arrow$ $h equals fraction numerator sec invisible function application theta over denominator square root of 2 end fraction comma k equals fraction numerator text cosec end text theta over denominator 2 end fraction rightwards double arrow cos invisible function application theta equals fraction numerator 1 over denominator h square root of 2 end fraction comma sin invisible function application theta equals fraction numerator 1 over denominator 2 k end fraction$
$rightwards double arrow$ $fraction numerator 1 over denominator 2 h to the power of 2 end exponent end fraction plus fraction numerator 1 over denominator 4 k to the power of 2 end exponent end fraction equals 1$ , \Required locus is $fraction numerator 1 over denominator 2 x to the power of 2 end exponent end fraction plus fraction numerator 1 over denominator 4 y to the power of 2 end exponent end fraction equals 1$ .
Trick: The locus of mid-points of the portion of tangents to the ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$ intercepted between axes is $a to the power of 2 end exponent y to the power of 2 end exponent plus b to the power of 2 end exponent x to the power of 2 end exponent equals 4 x to the power of 2 end exponent y to the power of 2 end exponent$
i.e., $fraction numerator a to the power of 2 end exponent over denominator 4 x to the power of 2 end exponent end fraction plus fraction numerator b to the power of 2 end exponent over denominator 4 y to the power of 2 end exponent end fraction equals 1$ or $fraction numerator 1 over denominator 2 x to the power of 2 end exponent end fraction plus fraction numerator 1 over denominator 4 y to the power of 2 end exponent end fraction equals 1$ .

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