Name: Orthogonal Curves
Uploaded: 2019-07-19
Duration: 11 min 48 s
Description: Have you heard the term 'orthogonal curves'? Can you find angles between two curves? Do watch this video and find all such answers!! Have fun!!

Question

Prove that the orthocentre of any triangle formed by three tangents to a parabola lies on the directrix.

Accepted Answer

Let the Equations of tangent to the parabola $y^{2} = 4 a x$ in slope form is,

$y = m_{1} x + \frac{a}{m_{1}} \dots (1)$

$y = m_{2} x + \frac{a}{m_{2}} \dots (2)$

$y = m_{3} x + \frac{a}{m_{3}} \dots (3)$

Orthocentre of the triangle formed by above 3 tangent is formed by the intersection of altitudes dram from each vertices of the triangle to its opposite side.

The line passing through point of intersection of equations $(2) and (3)$ is and perpendicular to $(1)$ is having

slope $= - \frac{1}{m_{1}}$ and equation is,

$y - a (\frac{1}{m_{2}} + \frac{1}{m_{3}}) = - \frac{1}{m_{1}} (x - \frac{a}{m_{2} m_{3}})$

$\Rightarrow y + \frac{x}{m_{1}} = a (\frac{1}{m_{2}} + \frac{1}{m_{3}} + \frac{a}{m_{1} m_{2} m_{3}}) \dots (4)$

Similarly, the Equation of line passing through point of intersection of equation $(3) and (1)$ is and perpendicular to $(2)$ is,

$\Rightarrow y + \frac{x}{m_{2}} = a (\frac{1}{m_{3}} + \frac{1}{m_{1}} + \frac{a}{m_{1} m_{2} m_{3}}) \dots (5)$

And through $(1) and (2)$ and perpendicular to $(3)$ is,

$\Rightarrow y + \frac{x}{m_{3}} = a (\frac{1}{m_{2}} + \frac{1}{m_{1}} + \frac{a}{m_{1} m_{2} m_{3}}) \dots (6)$

Now, the point of intersection of line $(4), (5) and (6)$ is the required Orthocentre,

$x = - a$

$y = a (\frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}} + \frac{1}{m_{1} m_{2} m_{3}})$

And Equation of Directrix of Parabola is also $x = - a$ .

Therefore, we can say that Orthocentre lies on Directrix.

Prove that the orthocentre of any triangle formed by three tangents to a parabola lies on the directrix.

Important Questions on Parabola

Learn and Prepare for any exam you want !