Question

The value of λ, for which the line 2x - $\frac{8}{3}$λy = -3 is a normal to the conic x² + $\frac{y^{2}}{4}$ = 1 is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C.
  • $\frac{\sqrt{3}}{2}$
• D. $\frac{3}{8}$

Answer 1

Answer: (D)

$\frac{3}{8}$

Answer 2

We know that the equation of the normal at point (a cos θ, b sin θ) on the curve x² + $\frac{y^{2}}{4}$ = 1 is given by ax sin θ - b y cosec θ = a² – b² ...(i)

Comparing equation (I) with 2x - $\frac{8}{3}$λy = -3.We get,

a sin θ = 2, b cosec θ$\frac{8}{3}$λ or ab = $\frac{16}{3}$λ ...(ii)

$\because$a = 1, b = 2, ∴ 2 = $\frac{16}{3}$λ or λ = 3/8.