Question

Find the condition on $a$ and $b$ so that the two tangents drawn to the parabola $y^{2} = 4ax$ from a point are normal to the parabola $x^2 = 4by$.
A) $a^{2} > 8b^{2}$
B) $4a^{2} > b^{2}$
C) $a^{2} > 4b^{2}$
D) $8a^{2} > b^{2}$

Answer 1

The equation of tangent to the parabola $y^{2} = 4ax$ is given by $y = mx+\dfrac{a}{m}$. Here $m$ is the slope of the line.
The equation of a line passing through the point $(x_{1},y_{1})$ having slope $m$ is given as,
$y-y_{1} = m(x-x_{1})$.

Complete step-by-step answer:
Given the equation of parabolas are $y^{2} = 4ax$ and $x^2 = 4by$.
Consider the parabola $x^2 = 4by$.
Any point on this parabola has the coordinates given by $(2bt,bt^{2})$.
Differentiate the equation on both sides with respect to $x$ to find the slope of the tangent to the parabola.
\begin{align*}2x &= 4b\dfrac{dy}{dx}\\\ \dfrac{dy}{dx} &= \dfrac{x}{2b}\end{align*}
Since the normal is perpendicular to the tangent at a point, slope of the normal is given as,
\begin{align*}m_{N} &= -\dfrac{1}{dy/dx}\\\ &= -\dfrac{2b}{x}\end{align*}
Thus the slope of the normal at the point $(2bt,bt^{2})$ is $-\dfrac{2b}{2bt} = -\dfrac{1}{t}$.
And the equation of the normal at that point is given using the formula $y-y_{1} = m(x-x_{1})$ as,
\begin{align*}(y-bt^{2}) &= -\dfrac{1}{t}(x-2bt)\\\ y &= -\dfrac{x}{t}+2b+bt^{2}\end{align*}
Since the tangents drawn to the parabola $y^{2} = 4ax$ from a point are normal to the parabola $x^{2} = 4by$, the equation $y = -\dfrac{x}{t}+2b+bt^{2}$ is also the equation of the tangent.
But the equation of the tangent drawn to the parabola $y^{2} = 4ax$ is given using the formula $y = mx+\dfrac{a}{m}$.
This implies,
\begin{align*}2b+bt^{2} &= \dfrac{a}{-1/t}\\\ bt^{2}-at+2b &= 0\end{align*}
The above quadratic equation has real roots and so its discriminant is greater than 0, i.e.,
\begin{align*}a^{2}-4(2b)\cdot b &> 0\\\ a^{2}-8b^{2} &> 0\\\ a^{2} &> 8b^{2}\end{align*}

Hence option (A) is correct.

Note: The condition can also be determined by determining the equation of the tangent line to the parabola ${y^2} = 4ax$ and the equation of normal to the parabola ${x^2} = 4by$ and then equating both the equations.