10-2. Parabola, Ellipse, Hyperbola
hard

वत्त $x^{2}+y^{2}=25$ की उस जीवा, जो अति परवलय $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ की स्पर्श रेखा है, के मध्य बिंदु का बिंदुपथ है

A

$\left(x^{2}+y^{2}\right)^{2}-16 x^{2}+9 y^{2}=0$

B

$\left(x^{2}+y^{2}\right)^{2}-9 x^{2}+144 y^{2}=0$

C

$\left(x^{2}+y^{2}\right)^{2}-9 x^{2}-16 y^{2}=0$

D

$\left(x^{2}+y^{2}\right)^{2}-9 x^{2}+16 y^{2}=0$

(JEE MAIN-2021)

Solution

$y-k=-\frac{h}{k}(x-h)$

$ky – k ^{2}=- hx + h ^{2}$

$hx + ky = h ^{2}+ k ^{2}$

$y =-\frac{ hx }{ k }+\frac{ h ^{2}+ k ^{2}}{ k }$

tangent to $\frac{ x ^{2}}{9}-\frac{ y ^{2}}{16}=1$

$c ^{2}= a ^{2} m ^{2}- b ^{2}$

$\left(\frac{ h ^{2}+ k ^{2}}{ k }\right)^{2}=9\left(-\frac{ h }{ k }\right)^{2}-16$

$\left( x ^{2}+ y ^{2}\right)^{2}=9 x ^{2}-16 y ^{2}$

Standard 11
Mathematics

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