Conics Chord of Contact (1 Viewer)

Ambility · Mar 30, 2016

Can someone show how you would set out working for this question?

$\\$Find the equation of the chord of contact of the tangents to $x^2-16y^2=16$ from the point $(2,-3)$.$$

InteGrand · Mar 30, 2016

Ambility said:
Can someone show how you would set out working for this question?

$\\$Find the equation of the chord of contact of the tangents to $x^2-16y^2=16$ from the point $(2,-3)$.$$

$\noindent The hyperbola is: $\frac{x^2}{4^2}-y^2 = 1$. Now, use the fact that the equation of the chord of contact from the point $\left(x_0,y_0\right)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{x_0 x}{a^2}-\frac{y_0y}{b^2}=1$, assuming such a chord exists (which'll happen iff the point $\left(x_0,y_0\right)$ isn't `inside' one of the hyperbola's branches. This is because if the point is inside, no tangents exists in the first place, whereas if it is outside, then there do exist do two tangents; if the point is on the ellipse itself, they wouldn't ask for a `chord of contact', but rather just a tangent, which'd have the equation given above though anyway. As an exercise one may wish to prove these facts about the existence of tangents depending on whether the point is inside or outside the hyperbola's branches.).$

$\noindent (The proof for the equation of the chord of contact is usually found in HSC 4U textbooks.)$

Ambility · Mar 30, 2016

In a test, would we be allowed to assume the identity $\frac{x_0x}{a^2}-\frac{y_0y}{b^2}=1$ for chords of contact, or would we have to prove this? If a proof is required, how would we prove it?

InteGrand · Mar 30, 2016

Ambility said:
In a test, would we be allowed to assume the identity $\frac{x_0x}{a^2}-\frac{y_0y}{b^2}=1$ for chords of contact, or would we have to prove this? If a proof is required, how would we prove it?

I think it needs to be proved. Check your HSC 4U textbook for the proof (it should be in both the Patel and Arnold & Arnold books). Alternatively, you can check it here from this old thread, which does it for the ellipse:

funnytomato said:
I just uploaded it

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The proof for the hyperbola is basically identical, just replace b^2 with –b^2 everywhere. (This proof assumes the result for the equation of a tangent at a given point on the conic; this result can be proved using calculus and point-gradient form of a line.)

Ambility · Mar 30, 2016

InteGrand said:
I think it needs to be proved. Check your HSC 4U textbook for the proof (it should be in both the Patel and Arnold & Arnold books). Alternatively, you can check it here from this old thread, which does it for the ellipse:

.

The proof for the hyperbola is basically identical, just replace b^2 with –b^2 everywhere. (This proof assumes the result for the equation of a tangent at a given point on the conic; this result can be proved using calculus and point-gradient form of a line.)

Cheers.

Conics Chord of Contact (1 Viewer)

Ambility

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InteGrand

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Ambility

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InteGrand

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Ambility

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