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Find the points of intersection of two perpendicular tangents to a parabola $y^2=4ax$.

Any point on the parabola is $(at^2,2at)$. Hence if they intersect at $(x_1,y_1)$ then $\dfrac{2at-y_1}{at^2-x_1}=\dfrac{1}{t}\implies at^2-ty_1+x_1=0$ since the slope at $(at^2,2at)=\dfrac{1}{t}$.

In order to the tangents to perpendicular $t_1=t_2$.

But how to find the points of intersection between them.Please help

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  • http://www.askiitians.com/iit-jee-coordinate-geometry/tangent-to-a-parabola.aspx#point-of-intersection-of-tangents – lab bhattacharjee Jan 11 '18 at 11:14
  • See if this helps you : https://math.stackexchange.com/questions/1602586/finding-the-locus-of-the-points-of-intersection-of-tangents-to-a-parabola –  Jan 11 '18 at 11:14
  • An easy way to find the intersection of two tangents is via pole-polar relationships: the intersection is the pole of the chord of contact. – amd Jan 11 '18 at 23:56

3 Answers3

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Let $(x,y)=(at^{2},2at)$, then the equation of the tangent at $[t]$ is

$$x-ty+at^{2}=0 \tag{1}$$

If tangents at $[t]$ and $[t']$ are perpendicular, $$tt'=-1$$

Then the equation of another tangent at $[t']$ is $$x+\frac{y}{t}+\frac{a}{t^{2}}=0$$

That is $$t^{2}x+ty+a=0 \tag{2}$$

$(1)+(2)$, $$(1+t^2)x+a(t^2+1)=0$$

$$\fbox{$x=-a$}$$

The locus is the directrix of the parabola which is the limiting case for director circle of an ellipse or a hyperbola.

Useful fact:

Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x',y')$ is given by

$$ax'x+h(y'x+x'y)+by'y+g(x+x')+f(y+y')+c=0$$

Also see another answer of mine for your further interest.

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Ng Chung Tak
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From this, $$m^2x-my+a=0$$

If $m_1,m_2$ are the roots of the above equation, $m_1+m_2=?, m_1m_2=?$

If the angle between the tangents is $\theta$ $$\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$$

Square both sides and use $$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2$$

lab bhattacharjee
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Here’s a somewhat different approach. Working in homogeneous coordinates, if we represent a non-degenerate conic with a matrix $C$, i.e., the conic is the set $\{\mathbf x \mid \mathbf x^TC\mathbf x=0 \}$, then every tangent line $\mathbf l$ to the parabola satisfies the dual conic equation $\mathbf l^TC^{-1}\mathbf l=0$. For this parabola, $$C = \begin{bmatrix}0&0&2a\\0&-1&0\\2a&0&0\end{bmatrix},$$ so by setting $\mathbf l = [\lambda:\mu:\tau]$ and solving for $\tau$, we find that its tangents are of the form $\mathbf l = \left[\lambda:\mu:\frac{a\mu^2}\lambda\right]$, $\lambda\ne0$. (If $\lambda=0$, the line is horizontal and can’t be tangent to the parabola.) The perpendicular tangent to this can be found by making the substitutions $\lambda\mapsto\mu$ and $\mu\mapsto-\lambda$ to get $\mathbf m = \left[\mu:-\lambda:\frac{a\lambda^2}\mu\right]$ ($\mu\ne0$), and the intersection of these two lines is $$\mathbf l\times\mathbf m = \left[a(\lambda^2+\mu^2) : a\left(\frac{\mu^3}\lambda-\frac{\lambda^3}\mu\right) : -(\lambda^2+\mu^2)\right]$$ which is $\left(-a, a\left(\frac\lambda\mu-\frac\mu\lambda\right)\right)$ in inhomogeneous Cartesian coordinates. The parameters $\lambda$ and $\mu$ can be chosen to produce any value whatsoever for the $y$-coordinate, so the locus of intersection is the line $x=-a$, the parabola’s directrix.

amd
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