Question

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

• A. $a^2+b^2$
• B. $\sqrt{(a^2+b^2)}$
• C. $(a^2+b^2{)}^2$
• D. $Noneofthese$

Answer 1

The correct option is B $\sqrt{(a^2+b^2)}$

Ellipse:$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1---\left(i\right)$

Tangent at any point on $\left(i\right)$ be $y=mx+\sqrt{a^2{m}^2+b^2}$

Intercept form$\Rightarrow \frac{y}{\sqrt{a^2{m}^2+b^2}}+\frac{x}{\frac{\sqrt{a^2{m}^2+b^2}}{(-m)}}=1$

intercept at $x-$axis= that on $y-$axis

$\Rightarrow \frac{\sqrt{a^2{m}^2+b^2}}{-m}=\sqrt{a^2{m}^2+b^2}$

$\Rightarrow m=-1$

so intercept length $l=\sqrt{a^2{m}^2+b^2}$

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

$=\sqrt{a^2+b^2}$