The least intercept made by a tangent to the ellipse (x^2)/64+(y^2)/49=1 with coordinate axes is

Correct answer: (C) 15. Given, ellispe : (x^2)/64+(y^2)/49=1 Here, a^2=64 ⇒ a=8 b^2=49 ⇒ b=7 When intercept made by tangent on coordinate axes is minimum, then least value of intercept =a+b=15

Q108 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics

The least intercept made by a tangent to the ellipse $\frac{x^2}{64}+\frac{y^2}{49}=1$ with coordinate axes is

A
$40$
B
$10$
Correct answer
C
$15$
D
$100$

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 $=a+b=15$

Hence, the correct answer is option: 'C'.

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