Question

Find the value of '$\lambda$' for which $2x-y+\lambda =0$ touches the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$

Answer 1

$2x-y+\lambda =0$

$y=2x+\lambda$

$\frac{x^2}{25}+\frac{(2x+\lambda {)}^2}{9}=1$

$9x^2+25(4x^2+{\lambda }^2+4x\lambda )=225$

$9x^2+100x^2+25{\lambda }^2+100x\lambda =225$

$109x^2+100\lambda x+(25{\lambda }^2-225)=0$

For tangent, discriminant should be o,

$\therefore (100\lambda {)}^2-4(2{\lambda }^2-225)109=0$

$10000{\lambda }^2-10900{\lambda }^2+98100=0$

$900{\lambda }^2=98100$

$\lambda =\pm \sqrt{109}$