Question

lf the curves $\frac{x^2}{a^2}+\frac{y^2}{4}=1$ and $y^2=16x$ intersect at right angles then value of $a^2$ is

• A. $2$
• B. $4$
• C. $\frac{8}{3}$
• D. $\frac{16}{3}$

Answer 1

Answer: (A)

$2$

$y^{\prime }=\frac{16}{2y}=m_1$

$y^{\prime }=-\frac{4x}{a^{2}y}=m_1$

$m_1{m}_2=-1$

$\frac{32x}{a^2{y}^2}=1$

$\frac{32x}{a^{2}16x}=1$

$a^2=2$