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The minimum & maximum distance of a point $(-1,6)$ from the ellipse $\mathrm{25 x^2+9 y^2+50 x+36 y-164=0}$ are l and m respectively then $\mathrm{m^2-l^2}$ equals
Option: 1
169

Option: 2
13

Option: 3
100

Option: 4
160

Answers (1)

best_answer

$\mathrm{\text { Let } S=25 x^2+9 y^2+50 x+36 y-164=0}$ $....(i)$

$\mathrm{\begin{aligned} \therefore \quad & S_{(-1,6)}=25(-1)^2+9(6)^2+50(-1)+36(6)-164 \\\\ & =25+324-50+216-164>0 \end{aligned}}$

which means the point (-1,6) is outside the ellipse.

Now, (i) can be rewritten as

$\mathrm{ \frac{(x+1)^2}{9}+\frac{(y+2)^2}{25}=1=\frac{X^2}{a^2}+\frac{Y^2}{b^2} }$
$\therefore \quad$ Centre of ellipse is $(-1,-2)$

& axis is parallel to y-axis as $\mathrm{b>a.}$

$\mathrm{\therefore \quad}$ Vertices are $\mathrm{x+1=0 \, \, \&}$

$\mathrm{ y+2= \pm 5 \text { i.e., }(-1,3) \text { and }(-1,-7) }$

The minimum distance $\mathrm{=l=R P=6-3 =3}$

units and the maximum distance

$\mathrm{=m=R Q=6-(-7)=13}$

units $\mathrm{\therefore \quad m^2-l^2=13^2-3^2=160}$

Posted by

Ritika Jonwal

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