Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $1$
• D. $\frac{4}{3}$

Answer 1

Answer: (D)

$\frac{4}{3}$

$y^{\prime }=\frac{16}{3y^2}=m_1$
$y^{\prime }=\frac{-4x}{y\times 4a^2}=m_2$
$m_1{m}_2=-1$
$\frac{16+x4}{3y^2\times y\times a^2}=1$
$\frac{16+x4}{3\times 16\times a^2}=1$
$a^2=\frac{4}{3}$