Question

Prove that two tangents to the parabolas \[{y^2} = 4a(x + a)\] and \[{y^2} = 4a'(x + a')\] which are at right angles to one another, meet on the straight line \[x + a + a' = 0\]. Show also that this straight line is the common chord of the two parabolas.

Answer 1

Hint: Use the equation of the tangent to the parabola \[{y^2} = 4a(x + a)\], that is, \[y = m(x + a) + \dfrac{a}{m}\] to determine the equation for the tangents to both the parabolas and use the formula for perpendicular lines to find the intersection of the lines. Then, to determine the equation of the common chord to the parabolas, subtract both equations and simplify.

Complete step-by-step answer:
We know that equation of tangent to the parabola \[{y^2} = 4a(x + a)\] is given as follows:
\[y = m(x + a) + \dfrac{a}{m}..............(1)\]
The equation of tangent to the parabola \[{y^2} = 4a'(x + a')\] is given as follows:
\[y = m'(x + a') + \dfrac{{a'}}{{m'}}..............(2)\]
Let the tangents intersect at (h, k), then we have the two equations:
\[k = m(h + a) + \dfrac{a}{m}..............(3)\]
\[k = m'(h + a') + \dfrac{{a'}}{{m'}}..............(4)\]
It is given that these lines are perpendicular, hence, the product of their slopes is equal to – 1.
\[mm' = - 1\]
\[m' = \dfrac{{ - 1}}{m}...........(5)\]
Substituting equation (5) in equation (4), we get:
 \[k = - \dfrac{1}{m}(h + a') - a'm..........(6)\]
Subtracting equation (6) from equation (3), we get:
\[k - k = m(h + a) + \dfrac{a}{m} + \dfrac{1}{m}(h + a') + a'm\]
Simplifying the equation, we get:
\[m(h + a + a') + \dfrac{1}{m}(h + a + a') = 0\]
Taking common terms, we get:
\[\left( {m + \dfrac{1}{m}} \right)(h + a + a') = 0\]
Hence, we get:
\[h + a + a' = 0\]
h is the x coordinate of the point of intersection. Hence, we can replace it with x to get the desired line on which the point of intersection lies.
\[x + a + a' = 0\]
To determine the equation of the common chord we subtract given two parabolas,.
\[{y^2} - {y^2} = 4a(x + a) - 4a'(x + a')\]
\[0 = 4ax + 4{a^2} - 4a'x - 4{a'^2}\]
Dividing both sides of the equation by 4, we get:
\[0 = ax + {a^2} - a'x - {a'^2}\]
Taking x as a common term, we get:
\[0 = x(a - a') + ({a^2} - {a'^2})\]
We know that \[{a^2} - {b^2} = (a + b)(a - b)\], then, we have:
\[x(a - a') + (a + a')(a - a') = 0\]
Simplifying, we get:
\[(a - a')(x + a + a') = 0\]
We know that \[a \ne a'\], then, we have:
\[x + a + a' = 0\]
Therefore, the common chord is the same as the line of intersection of the two tangents.

Note: The common chord is drawn between the two points of intersection of the two parabolas and hence, at this point the both parabola equations are satisfied. Hence, the equation is obtained by subtracting the equations of the parabolas. Students should remember the equation of tangent to the parabola i.e For parabola \[{y^2} = 4ax\] the equation of tangent is \[y = mx + \dfrac{a}{m}\]. In the above question x is replaced by x+a to get the required tangent equation.