If the segment intercepted by the parabola $y^2 =4ax$ with the line lx +my +n=0 subtends a right angle at the vertex, then

Question

Problem :

If the segment intercepted by the parabola $y^2 =4ax$ with the line lx +my +n=0 subtends a right angle at the vertex, then

(a) 4al +n=0

(b)4am +n=0

(c) al +n=0

(d) 4al +4am +n=0

My working :

Let the point of intersection and the parabola is A (a,b) and B (c,d) therefore it should satisy the equation of line : $\Rightarrow al +mb +n=0 .......(i) ; cl +md +n=0 ......(ii)$

Also the vertex of the parabola is at the origin O(0,0) and the point O(0,0) ; A(a,b) and O(0,0) ; B(c,d) are at right perpendicular to each other

$\Rightarrow \frac{b}{a} \times \frac{d}{c} = -1 ..........(iii)$

Also AOB forms a right angle triangle therefore : $AO^2 + BO^2 = AB^2$

Is this the right way of approaching this problem or is there some shorter method to solve this please suggest..thanks..

Answer 1

The parabola $y^2=4ax$ can be written parametrically as $x=at^2,y=2at$

Let $\displaystyle P(au^2,2au), Q(av^2,2av)$ be two intersections and $O(0,0)$ be the vertex with $u\cdot v\ne0$

As $\displaystyle PO\perp OQ,$ $$\frac{2au-0}{au^2-0}\cdot \frac{2av-0}{av^2-0}=-1\implies uv=-4\ \ \ \ (1)$$

Now $P,Q$ lies on the straight line $\displaystyle lx+my+n=0$

So, $\displaystyle l(au^2)+m(2au)+n=0\implies lau^2+2mau+n=0$ and $\displaystyle lav^2+2mav+n=0$

Clearly, $u,v$ are the roots of the Quadratic equation $\displaystyle lat^2+2mat+n=0$

$\displaystyle\implies uv=\frac n{la}\ \ \ \ (2)$

Compare $(1),(2)$

Just to remind the condition derived is only necessary, but not sufficient for perpendicularity/Orthogonality or $PO,OQ$

Answer 2

Vector solution: first I parametrize the parabola as $ \gamma(t)= (x=at^2,y=2at)$, then I take the parameter value for two point which tangent cut as $t_1$ and $t_2$, now it must be that:

$$ \gamma(t_2) - \gamma(t_1) \times < 1, - \frac{l}{m}>=0$$

Since the chord vector is parallel to the direction of the line. Now, it must be also be that that if we treat $\gamma(t)$ as a position vector from origin then:

$$ \gamma(t_1) \cdot \gamma(t_2)=0$$

Due to the orthogonality condition in the question, from the above two equations I get the two following conditions on the parameter value:

$$ t_1 t_2 = -4 \tag{1}$$

$$t_1 + t_2 = \frac{2m}{l} \tag{2}$$

Now, I plug cartesian parameterization i.e: $\frac{y^2}{4a}=x$ into the equation of line:

$$ l\frac{y^2}{4a}+ m y +n=0$$

Rearranging:

$$ y^2 + \frac{4a}{l} m y + \frac{4a}{l}n = 0 \tag{3}$$

Let $y_1$ and $y_2$ be the roots of the above equation, then by the definition of $y$ it must be that:

$$y_1 y_2= (2at_1)(2at_2)=-16a^2$$

But by vieta, the above expression must be equal to $\frac{4a}{l}n$, rearranging, we find that $ 4al+n=0$