Question

The pole of lx + my = 1 with respect to the ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ lies on the ellipse $\frac{x^{2}}{4a^{2}} + \frac{y^{2}}{4b^{2}} = 1$ if

• A. b² + a²m² = 4
• B. b² + a²m² = 2
• C. a² + b²m² = 4
• D. None of these

Answer 1

Answer: (B)

b² + a²m² = 2

Answer 2

Pole of lx + my = 1 w.r.t. the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is (a²l, b²m).

It will lie on the ellipse $\frac{x^{2}}{4a^{2}} + \frac{y^{2}}{4b^{2}} = 1$

if $\frac{a^{4}l^{2}}{4a^{2}} + \frac{b^{4}m^{2}}{4b^{2}} = 1$ or if a²l² + b²m² = 4, which is the required condition.