Question

Find the equations of the tangents of the parabola y2 = 12x, which passes through the point (2,5).

Answer 1

Answer on Question #56979 – Math – Analytic Geometry

Find the equations of the tangents of the parabola $y^{2} = 12x$, which passes through the point (2,5).

Solution

Method 1

The equation of the tangent is given by

$$y = y(x_0) + y'(x_0)(x - x_0)$$

If $y^2 = 12x$, then $y = \sqrt{12x}$ or $y = -\sqrt{12x}$.

**Case 1.** Assume $y = \sqrt{12x}$. Then

$$y' = \left(\sqrt{12x}\right)' = \sqrt{12} \left(x^{1/2}\right)' = \sqrt{12} \cdot \frac{1}{2} \cdot x^{-1/2} = \sqrt{\frac{3}{x}}$$ $$y(x_0) = \sqrt{12x_0}$$ $$y'(x_0) = \sqrt{\frac{3}{x_0}}, \quad x_0 \neq 0.$$

If $x_0 = 0$, then the derivative $y'$ is not finite at point $x_0 = 0$, so the equation of tangent is $x = 0$, but this tangent does not pass through the point (2,5), therefore we reject the case, where $x_0 = 0$. Thus, $x_0 \neq 0$.

So

$$y = \sqrt{12x_0} + \sqrt{\frac{3}{x_0}} (x - x_0).$$

Besides,

$$5 = \sqrt{12x_0} + \sqrt{\frac{3}{x_0}} (2 - x_0),$$

because the tangent line passes through the point (2,5),

Multiplying both sides by $x_0 \neq 0$,

$$5 \cdot \sqrt{x_0} = 2\sqrt{3} \cdot x_0 + \sqrt{3} (2 - x_0)$$ $$\sqrt{3} x_0 - 5 \sqrt{x_0} + 2\sqrt{3} = 0$$

Substitute $\sqrt{x_0} = t \geq 0$, then

$$\sqrt{3} t^2 - 5t + 2\sqrt{3} = 0$$ $$D = 25 - 4 \cdot \sqrt{3} \cdot 2\sqrt{3} = 25 - 24 = 1$$ $$t_1 = \frac{5 - 1}{2\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$ $$t_1 = \frac{5 + 1}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$$

Hence $x_0 = (t_1)^2 = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}$ or $x_0 = (t_2)^2 = (\sqrt{3})^2 = 3$.

Thus,

$$y = \sqrt {12 \cdot \frac {4}{3}} + \sqrt {\frac {3 \cdot 3}{4}} \left(x - \frac {4}{3}\right)$$

or

$$y = \sqrt {12 \cdot 3} + \sqrt {\frac {3}{3}} (x - 3).$$

They are equivalent to

$$y = 2 + \frac {3}{2} x$$

or

$$y = x + 3.$$

Case 2. Assume $y = -\sqrt{12x}$. Then

$$y' = - \left(\sqrt {12x}\right)' = - \sqrt {12} \left(x ^ {\frac {1}{2}}\right)' = - \sqrt {12} \cdot \frac {1}{2} \cdot x ^ {- \frac {1}{2}} = - \sqrt {\frac {3}{x}}$$ $$y (x _ {0}) = - \sqrt {12 x _ {0}}$$ $$y' (x _ {0}) = - \sqrt {\frac {3}{x _ {0}}}$$

So

$$y = - \sqrt {12 x _ {0}} - \sqrt {\frac {3}{x _ {0}}} (x - x _ {0})$$ $$5 = - \sqrt {12 x _ {0}} - \sqrt {\frac {3}{x _ {0}}} (2 - x _ {0}),$$

because the tangent line passes through the point (2,5).

Multiplying both sides by $x_0 \neq 0$,

$$5 \cdot \sqrt {x _ {0}} = - 2 \sqrt {3} \cdot x _ {0} - \sqrt {3} (2 - x _ {0})$$ $$- \sqrt {3} x _ {0} - 5 \sqrt {x _ {0}} - 2 \sqrt {3} = 0$$

Substituting $\sqrt {x _ {0}} = t \geq 0$, get

$$\sqrt {3} t ^ {2} + 5 t + 2 \sqrt {3} = 0,$$ $$D = 25 - 4 \cdot \sqrt {3} \cdot 2 \sqrt {3} = 25 - 24 = 1,$$ $$t _ {1} = \frac {- 5 - 1}{2 \sqrt {3}} = \frac {- 3}{\sqrt {3}} = - \sqrt {3} < 0,$$ $$t _ {1} = \frac {- 5 + 1}{2 \sqrt {3}} = \frac {- 2}{\sqrt {3}} < 0,$$

Hence this case does not satisfy assumption $\sqrt {x _ {0}} = t \geq 0$.

Method 2

Equation of tangent line of $y^{2} = 2px$ at point $(x_0, y_0)$ is $yy_0 = p(x + x_0)$ , where $y_0^2 = 2px_0$ .

It follows $x_0 = \frac{y_0^2}{12}$ from $y_0^2 = 12x_0$ .

In this problem $y^{2} = 12x$ , $p = 6$ and $(x,y) = (2,5)$ satisfies the equation $yy_{0} = p(x + x_{0})$ , that is,

$5y_{0} = 6(2 + x_{0})$

Substituting for $x_0 = \frac{y_0^2}{12}$ into $5y_0 = 6(2 + x_0)$ get $5y_0 = 6\left(2 + \frac{y_0^2}{12}\right)$ , hence

$5y_{0} = 6\frac{1}{12} (24 + y_{0}^{2}),$

$y_0^2 - 10y_0 + 24 = 0,$

$(y_0 - 6)(y_0 - 4) = 0,$

therefore $y_0 = 6$ or $y_0 = 4$ , hence

$x_0 = \frac{y_0^2}{12} = \frac{6^2}{12} = \frac{36}{12} = 3$

or

$x_0 = \frac{y_0^2}{12} = \frac{4^2}{12} = \frac{16}{12} = \frac{4}{3}$

Finally there exists two tangents:

substituting for $(x_0,y_0) = (3,6)$ into $yy_{0} = p(x + x_{0})$ get $6y = 6(x + 3)$ , that is, $y = x + 3$ ;

substituting for $(x_0,y_0) = \left(\frac{4}{3},4\right)$ into $yy_{0} = p(x + x_{0})$ get $4y = 6\left(x + \frac{4}{3}\right)$ , that is, $y = \frac{3}{2} x + 2$ .

Remark

The equations of the tangent of the parabola $y^{2} = 12x$ at the point, where $x = 2.5$ , are

$Y = \sqrt{30} +\sqrt{\frac{3}{2.5}} (x - 2.5)$

and

$$Y = - \sqrt {3 0} - \sqrt {\frac {3}{2 . 5}} (x - 2. 5)$$

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