The least intercept made by a tangent to the ellipse $\frac{x^2}{64}+\frac{y^2}{49}=1$ with coordinate axes is
The least intercept made by a tangent to the ellipse $\frac{x^2}{64}+\frac{y^2}{49}=1$ with coordinate axes is
$40$
$10$
$15$
$100$
Solution
Given, ellispe : $\frac{x^2}{64}+\frac{y^2}{49}=1$
Here, $a^2=64 \Rightarrow a=8$
$b^2=49 \Rightarrow b=7$
When intercept made by tangent on coordinate axes is minimum, then least value of intercept
$=a+b=15$