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A line $y = 2x+5$ intersects a hyperbola only at a point $(-2,1)$. The equation of one of its asymptotes is $3x+2y+1=0$. If hyperbola passes through the point $(-1,0)$ then what is the equation of hyperbola?

This is a question I found, and the solution has given the following equation as the equation of the hyperbola: $(2x-y+\lambda)(3x+2y+1)+k = 0$. Where did they get this equation from? I just know the basics of what an asymptote is, it is a line that touches the hyperbola at infinity, or basically a tangent at infinity. Where did they get this equation from?

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As line $y=2x+5$ intersects the hyperbola only at a single point (but it is not tangent), then this line must be parallel to an asymptote. It follows that the equation of the second asymptote must be $2x−y+λ=0$, for some $\lambda$.

On the other hand, if $2x−y+λ=0$ and $3x+2y+1=0$ are the equations of its asymptotes, then the equation of the hyperbola is $$ (2x−y+λ)(3x+2y+1)=k, $$ for some $k\ne0$. Substitute here the coordinates of the two given points to get the values of $\lambda$ and $k$.

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  • $\begingroup$ I'm so sorry to trouble you, but why is y=2x+5 not a tangent? $\endgroup$ – Hema May 20 '18 at 10:49
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    $\begingroup$ Because the text says "intersects" instead of "touches". $\endgroup$ – Intelligenti pauca May 20 '18 at 11:42
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    $\begingroup$ But it does says. ''...it intersect only at...'' $\endgroup$ – Aqua May 20 '18 at 11:44

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