Question

The value of $a^2$ if the curves $\frac{x^2}{a^2}+\frac{y^2}{4}=1$ and $y^3=16x$ cut orthogonally is

• A. $3/4$
• B. $1$
• C. $4/3$
• D. $4$

Answer 1

The correct option is C $4/3$

$l_1=\frac{x^2}{a^2}+\frac{y^2}{4}=1$
$l_2=y^3=16x$
$l_1$ has a slope $n_1$
$l_2$ has a slope $n_2$
$n_1{n}_2=-1$
$\frac{x^2}{a^2}+\frac{y^2}{4}=1$
$\frac{2x}{a^2}+\frac{1}{4}\times \frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{-x\times y}{a^2\times y}$
$n_1=\frac{-4x}{a^{2}y}$
$y^3=16x$
$3y^3\frac{dy}{dx}=16$
$\frac{dy}{dx}=\frac{16}{3y^2}$
$\therefore n_1{n}_2=-1$
$\frac{-4x}{a^{2}y}\left(\frac{16}{3y^2}\right)=-1$
$\frac{16\times 4x}{3a^2{y}^3}=1$
$\frac{16\times 4x}{3a^{2}16x}=1$
$a^2=4/3$