Question

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

Answer 1

Let the point of intersection of tangents be $P(h,k)$

Equation of tangent to $y^2=4a(x+a).....\left(i\right)$ is

$y=m(x+a)+\frac{a}{m}k=m(h+a)+\frac{a}{m}....\left(ii\right)$

Equation of tangent to $y^2=4a^{\prime }(x+a^{\prime }).......\left(iii\right)$ is

$y=m^{\prime }(x+a^{\prime })+\frac{a^{\prime }}{m^{\prime }}$

As both the lines are perpendicular

$\therefore$ $m{m}^{\prime }=-1$

$m^{\prime }=\frac{-1}{m}y=\frac{-1}{m}(x+a^{\prime })-m{a}^{\prime }k=\frac{-1}{m}(h+a^{\prime })-m{a}^{\prime }....\left(iv\right)$

Subtracting $\left(ii\right)$ and $\left(iv\right)$

$k-k=m(h+a)+\frac{a}{m}+\frac{1}{m}(h+a^{\prime })+m{a}^{\prime }mh+am+a^{\prime }m+\frac{1}{m}(h+a+a^{\prime })=0m(h+a+a^{\prime })+\frac{1}{m}(h+a+a^{\prime })=0(h+a+a^{\prime })(m+\frac{1}{m})=0(h+a+a^{\prime })=0$

Replacing $h$ by $x$

$x+a+a^{\prime }=0$

Hence proved.

Equation of common chord is obtained by subtracting both curves

$\therefore$ subtracting $\left(i\right)$ from $\left(iii\right)$

$y^2-y^2=4a^{\prime }(x+a^{\prime })-4a(x+a)0=4a^{\prime }x+4{a^{\prime }}^2-4ax-4a^2(a^{\prime }-a)x+{a^{\prime }}^2-a^2=0(a^{\prime }-a)x+(a^{\prime }-a)(a+a^{\prime })=0(a-a^{\prime })(x+a+a^{\prime })=0\Rightarrow x+a+a^{\prime }=0$

Hence proved that both the equations are same