Question

Equation of the common tangent to the parabola $y^2=4ax$ and $x^2=4by$ is

• A. $a^{\frac{1}{3}}x+b^{\frac{1}{3}}y+(ab{)}^{\frac{2}{3}}=0$
• B. $a^{\frac{1}{3}}x+b^{\frac{1}{3}}y=(ab{)}^{\frac{1}{3}}$
• C. $a^{\frac{1}{3}}x-b^{\frac{1}{3}}y+(ab{)}^{\frac{2}{3}}=0$
• D. $a^{\frac{1}{3}}x+b^{\frac{1}{3}}y-(ab{)}^{\frac{2}{3}}=0$

Answer 1

The correct option is A $a^{\frac{1}{3}}x+b^{\frac{1}{3}}y+(ab{)}^{\frac{2}{3}}=0$

For $y^2=4ax$ tangent is $y=mx+\frac{a}{m}\dots \left(1\right)$
For $x^2=4by$ tangent is $y=mx-b{m}^2\cdots \left(2\right)$
Since $\left(1\right)\&\left(2\right)$ represent the same tangent so,
$\frac{a}{m}=-b{m}^2$
$m=-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}$
Now, by $\left(2\right)$
$y=-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}x-b{\left(\frac{a}{b}\right)}^{\frac{2}{3}}$
$\Rightarrow a^{\frac{1}{3}}x+b^{\frac{1}{3}}y+(ab{)}^{\frac{2}{3}}=0$