Question

The equation of common tangent to the parabola $y^2=4ax$ and $x^2=4bx$ is

• A. ${xa}^{1/3}+{yb}^{1/3}+{\left(ab\right)}^{2/3}=0$
• B. $\frac{1}{a^{1/3}}+\frac{y}{b^{1/3}}+\frac{1}{{\left(ab\right)}^{2/3}}=0$
• C. ${xb}^{\frac{1}{3}}+{ya}^{\frac{1}{3}}-{\left(ab\right)}^{\frac{2}{3}}=0$
• D. $\frac{1}{b^{1/3}}+\frac{y}{a^{1/3}}-\frac{1}{{\left(ab\right)}^{2/3}}=0$

Answer 1

The correct option is A ${xa}^{1/3}+{yb}^{1/3}+{\left(ab\right)}^{2/3}=0$

let the common tangent be $y=mx+c\to \left(i\right)$

$y^2=4ax\to \left(ii\right)$

from (i) and (ii), point of intersect

$\begin{array}{cc}{m}^2{x}^2+x\left(2mc-4a\right)x+c^2=0& \\ D=0& \left\{\because itistangent\right\}\\ b^2-4ac=0\to \left(4\right)& \\ from\left(3\right)and\left(4\right)& \\ 16a^2-16amc=0& \\ a=mc\to \left(5\right)& \\ similarly,& \\ b{m}^2+c=0\to \left(6\right)& \\ from\left(5\right)and\left(6\right),weget& \\ \frac{b{a}^2}{c^2}+c=0& \\ \Rightarrow c=-{\left(b{a}^2\right)}^{\frac{1}{3}}\to \left(7\right)& \\ from\left(5\right)and\left(7\right)& \\ m=-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}\to \left(8\right)& \\ \Rightarrow equationof\tan gent& \\ isy+{\left(\frac{a}{b}\right)}^{\frac{1}{3}}x+{\left(b{a}^2\right)}^{\frac{1}{3}}=0& \\ or& \\ y{\left(b\right)}^{\frac{1}{3}}+{\left(a\right)}^{\frac{1}{3}}x+{\left(a^2{b}^2\right)}^{\frac{1}{3}}=0& \end{array}$