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I have the following parabola $$ P:y^2 − 6x − 6y + 3 = 0.$$ How can I find the tangent parallel to line $\ell: 3x − 2y + 7 = 0$?

I wouldn't have any problem with this problem if there was only one variable but how does this work with 2? Do I have to divert the parabola twice?

Robert Z
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xXx
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  • Read here https://math.stackexchange.com/questions/344768/tangent-line-to-a-general-conic-at-a-point?rq=1 and make a try! – Robert Z Dec 03 '18 at 13:06

3 Answers3

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$P:y^2 − 6x − 6y + 3 = 0, \ell: 3x − 2y + 7 = 0$

Note that these are still curves in the $xy$ plane, just not given in the conventional $y=f(x)$ form. This is called the implicit expression of a curve.

The slope of the tangent should be $\frac32$. Differentiate the equation of $P$ with respect to $x$,

$2yy'-6-6y'=0\implies y'=\frac3{y-3}=\frac32\implies y-3=2$ or $y=5$

$y^2-6x-6y+3=0\implies 25-6x-30+3=0\implies x=-\frac13$

The required tangent passes through $(-\frac13,5)$. Its equation is $y-5=\frac32(x+\frac13)\implies y=\frac32x+\frac{11}2$

Shubham Johri
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So the slope of a line is $3/2$ so a tangenthas equation $y=3x/2 +n$.

Since $3x = 2y-2n$ we have quadratic equation on $y$ with parameter $$y^2-6y+3 -2(2y-2n)=0$$ or $$y^2-10y+3 +4n=0$$ since it must have only one solution, a discriminant is 0:

$$ 100-4(3+4n)=0\implies n=11/2$$

so the equation of tangent is $\boxed{y={3\over 2}x +{11\over 2}}$.

nonuser
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The tangent line to this parabola at a point $(x_0,y_0)$ on it is given by the equation $$yy_0-3(x+x_0)-3(y+y_0)+3=0. \tag 1$$ (There’s a simple substitution rule to obtain the tangent to any conic given its implicit Cartesian equation.) Rearrange this equation into $$-3x+(y_0-3)y-3(x_0+y_0+3)=0.$$ Comparing its coefficients to those of the parallel line equation we have $y_0-3=2$. Substitute the resulting value of $y_0$ into the parabola’s equation and solve for $x_0$, then substitute both into equation (1).

amd
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