Question

A tangent is draw to the ellipse $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1$so that per

intercepted by the axes is minimum. The length of this part

of the tangent is

• A. 11
• B. 8
• C. 9
• D. ) 10

Answer 1

Answer: (C)

9

Answer 2

Equation of tangent $\frac { x } { 5 } \cos \phi + \frac { y } { 4 } \sin \phi = 1$

Length of AB = $\sqrt { \frac { 25 } { \cos ^ { 2 } \phi } + \frac { 16 } { \sin ^ { 2 } \phi } } = \sqrt { t }$

t = $\frac { 25 } { \cos ^ { 2 } \phi } + \frac { 16 } { \sin ^ { 2 } \phi } = 25 \sec ^ { 2 } \phi + 16 \operatorname { cosec } { } ^ { 2 } \phi$

$\frac { d t } { d \phi } = 50 \sec ^ { 2 } \phi \cdot \tan \phi - 32 \operatorname { cosec } { } ^ { 2 } \phi \cot \phi = 0$ $\tan ^ { 4 } \phi = \frac { 32 } { 50 }$⇒ $\tan ^ { 2 } \phi = \frac { 4 } { 5 }$

so $t = 25 \sec ^ { 2 } \phi + 16 \operatorname { cosec } { } ^ { 2 } \phi$

=$25 \cdot \frac { 9 } { 5 } + 16 \cdot \frac { 9 } { 4 }$ = 45 + 36

$\therefore$ t = 81 $\sqrt { t } = \sqrt { 81 } = 9$

so minimum length intercepted = '9'

so 'C' is correct.