Chord of Contact (1 Viewer)

nrumble42 · Dec 21, 2016

From what external point are the tangents to the ellipse x^2/8 + y^2/6 = 1 are drawn so that x-y+2 =0 is the chord of contact?

InteGrand · Dec 21, 2016

nrumble42 said:
From what external point are the tangents to the ellipse x^2/8 + y^2/6 = 1 are drawn so that x-y+2 =0 is the chord of contact?

$\noindent Let the desired external point be $\left(x_{0}, y_{0}\right)$. The chord of contact is $\frac{x_{0}x}{8} + \frac{y_{0}y}{6} = 1 \iff 3x_{0}x + 4y_{0}y = 24$. The desired chord of contact is $x-y = -2\iff -12x + 12y = 24$. So $3x_{0} = -12$, i.e. $x_{0} = -4$, and $4y_{0} = 12$, i.e. $y_{0} = 3$. So the desired external point is $\left(-4,3\right)$.$

nrumble42 · Dec 21, 2016

Why did you times by -12? Like, where did that come from?

InteGrand · Dec 21, 2016

nrumble42 said:
Why did you times by -12? Like, where did that come from?

$\noindent I just did that so that the RHS would be $24$ like it was in the equation $3x_{0}x + 4y_{0}y = 24$, so we could compare coefficients of $x$ and $y$.$

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Chord of Contact (1 Viewer)

nrumble42

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InteGrand

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nrumble42

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InteGrand

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