Question

If the curves $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { 4 } = 1$and y²⁼ 16x intersect at right angle then

• A. a = ±1
• B. a = ±$\sqrt{3}$
• C. a = ±$\frac{1}{\sqrt{3}}$
• D. ) a = ±$\sqrt{2}$

Answer 1

)

Sol. Clearly the given curves intersect for.

a ∈ R ~ {0}. For the parabola, $\frac { d y } { d x } = - \frac { 4 x } { a ^ { 2 } y }$. For the curves to intersect orthogonally,

=s>a²y² = 32x :z>17a²x = 32x => a² = 2