Question

If $3x+4y=12$ intersect the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at $P$ and $Q$, then find the point of intersection of tangents at $P$ and $Q.$

• A. $(\frac{25}{4},\frac{16}{3})$
• B. $(\frac{16}{3},\frac{25}{4})$
• C. $(\frac{-25}{4},\frac{16}{3})$
• D. $(\frac{-16}{3},\frac{-25}{4})$

Answer 1

The correct option is A $(\frac{25}{4},\frac{16}{3})$

let $(h,k)$ be the point of the contact of tangents
equation of chord of contact $\frac{hx}{25}+\frac{ky}{16}=1$
Comparing with $3x+4y=12$ : $\frac{h}{75}=\frac{k}{64}=\frac{1}{12}$
we get $(h,k)=(\frac{25}{4},\frac{16}{3})$