Question

If any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ intercepts equal length $1cm$ the axes, then $l=$

• A. $\sqrt{a^2+b^2}$
• B. $a^2+b^2$
• C. $(a^2+b^2{)}^2$
• D. None of these

Answer 1

The correct option is A $\sqrt{a^2+b^2}$

Condition for a line $y=mx+c$ to be tangent on ellipse is:

$c^2=a^2{m}^2+b^2$ ...(i)

now $y=mx+c$ can be written in intercept form as

$\frac{y}{c}+\frac{x}{-c/m}=1$

intercept has equal lengths

$c=-c/m$

$m=-1$

put it in eq (i)

$c^2=a^2(-1{)}^2+b^2$

$c^2=a^2+b^2$

$c=\sqrt{a^2+b^2}$ Ans.