Question

Prove that the parabolas $y^2-=4ax$ and $x^2=4by$ cut one another at an angle ${\tan }^{-1}\left(\frac{3a^{\frac{1}{3}}{b}^{\frac{1}{3}}}{2(a^{\frac{2}{3}}+b^{\frac{2}{3}})}\right)$

Answer 1

Let the common point be $C(p,q)$

To find the cutting angle we need to find the slopes of tangents at the common point

To find slopes, we need to differentiate with respect to $x$ and substitute the point in it.

Let $y^2=4ax$ ...... $\left(i\right)$ and $x^2=4by$ ....... $\left(ii\right)$

Differentiating $\left(i\right)$ with respect to $x,$ we get

$\frac{dy}{dx}=\frac{2a}{y}$ and the slope is $\frac{2a}{q}$

Similarly, differentiating $\left(ii\right)$ with respect to $x,$ we get

$\frac{dy}{dx}=\frac{x}{2b}$ and the slope is $\frac{p}{2b}$

Now, we have to find $C(p,q)$ in terms of $a$ and $b$

Solving$q^2=4ap$ and $p^2=4bq$

i.e. putting $q=\frac{p^2}{4b}$ in $q^2=4ap$, we get

$p=4a^{1/3}{b}^{2/3}$

Similarly, we get $q=4b^{1/3}{a}^{2/3}$

Now substituting in the slopes, we get

$m_1=\frac{2a^{1/3}}{b^{1/3}}$ and $m_2=\frac{a^{1/3}}{2b^{1/3}}$

Angle between tangents is ${\tan }^{-1}\left(\frac{m_1-m_2}{1+m_1{m}_2}\right)$

Substituting the values of slopes $m_1$ and $m_2$ in the above equation, we get

$\theta ={\tan }^{-1}\left(\frac{3a^{1/3}{b}^{1/3}}{2(a^{2/3}+b^{2/3})}\right)$