The line y = mx + c is a normal to the ellips | vedclass

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(c) As we know that the line $lx + my + n = 0$ is normal to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$,

if $\frac{{{a^2}}}{{{l^2}}} +

Vedclass · Accepted Answer

$ - \frac{{({a^2} - {b^2})m}}{{\sqrt {{a^2} + {b^2}{m^2}} }}$

Vedclass · Answer

(c) As we know that the line $lx + my + n = 0$ is normal to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$,

if $\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$.

But in this condition, we have to replace $l$ by $m$, $m$ by $-1$ and $n$ by $c$,

then the required condition is $c = \pm \frac{{({a^2} - {b^2})m}}{{\sqrt {{a^2} + {b^2}{m^2}} }}$.

The line $y = mx + c$ is a normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$, if $c = $

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