The line lx + my - n = 0 will be tangent to t | vedclass

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Question

(a) $y = \frac{{ - l}}{m}x + \frac{n}{m}$ is tangent to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, if

$\frac{n}{m} = \pm \sqrt {{b^2}

Vedclass · Accepted Answer

${a^2}{l^2} + {b^2}{m^2} = {n^2}$

Vedclass · Answer

(a) $y = \frac{{ - l}}{m}x + \frac{n}{m}$ is tangent to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, if

$\frac{n}{m} = \pm \sqrt {{b^2} + {a^2}{{\left( {\frac{l}{m}} \right)}^2}} $ or ${n^2} = {m^2}{b^2} + {l^2}{a^2}$.

The line $lx + my - n = 0$ will be tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, if

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