Question

Find the length of tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(3,5)$.

• A. $\sqrt{19}$
• B. $\sqrt{25}$
• C. $\sqrt{34}$
• D. $\sqrt{17}$

Answer 1

Answer: (C)

$\sqrt{34}$

Length of tangent from an external point (3,5) to

$E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$\Rightarrow \sqrt{E_{(3,5)}}$

$E_{(3,5)}=\frac{9}{a^2}+\frac{25}{b^2}-1$

$E_{(3,5)}=9b^2+25a^2-a^2{b}^2$

length of tangent $=\sqrt{9b^2+25a^2-a^2{b}^2}$

$=\sqrt{9+25}$

$=\sqrt{34}$