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Maths-

General

Easy

Question

Normal to the ellipse $fraction numerator x to the power of 2 end exponent over denominator 84 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 49 end fraction equals 1$ intersects the major and minor axis at P and Q respectively then locus of the point dividing segment PQ in 2 : 1 is -

$\frac{64 x^{2}}{25} + \frac{49 y^{2}}{100} = 1$
$\frac{64 x^{2}}{100} + \frac{49 y^{2}}{25} = 1$
$64 x^{2} + 49 y^{2} = 225$
$\frac{49 x^{2}}{100} + \frac{64 y^{2}}{25} = 1$

The correct answer is: $fraction numerator 64 x to the power of 2 end exponent over denominator 25 end fraction plus fraction numerator 49 y to the power of 2 end exponent over denominator 100 end fraction equals 1$

Equation of normal 8x sec $theta$ – 7y cosec $theta$ = 15
$P open parentheses fraction numerator 15 over denominator 8 end fraction cos invisible function application theta comma 0 close parentheses$ , Q $open parentheses 0 comma fraction numerator negative 15 over denominator 7 end fraction sin invisible function application theta close parentheses$

3h = $fraction numerator 15 over denominator 8 end fraction cos invisible function application theta$ , 3k = $negative fraction numerator 30 over denominator 7 end fraction sin invisible function application theta$
cos $theta$ = $fraction numerator 8 h over denominator 5 end fraction$ , sin $theta$ = $fraction numerator negative 7 k over denominator 10 end fraction$
hence locus $fraction numerator 64 x to the power of 2 end exponent over denominator 25 end fraction plus fraction numerator 49 y to the power of 2 end exponent over denominator 100 end fraction$ = 1

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