Question

If a line $3x-y=2$ intersects ellipse $\frac{x^2}{8}+\frac{y^2}{4}=1$ at points $A$ & $B$, then find co-ordinates of point of intersection of tangents at points $A$ & $B.$

• A. $(12,-2)$
• B. $(12,2)$
• C. $(-12,-2)$
• D. $(-12,2)$

Answer 1

Answer: (A)

$(12,-2)$

let $(h,k)$ be the pole
Then equation is $\frac{hx}{8}+\frac{ky}{4}=1$
comparing with $3x-y=2$, we get: $\frac{h}{24}=\frac{-k}{4}=\frac{1}{2}$
Therefore, $(h,k)=(12,-2)$