Conic Sections Question 268
Question: On the ellipse $ 4x^{2}+9y^{2}=1 $ , the points at which the tangents are parallel to the line $ 8x=9y $ are
[IIT 1999]
Options:
A) $ ( \frac{2}{5},\ \frac{1}{5} ) $
B) $ ( -\frac{2}{5},\ \frac{1}{5} ) $
C) $ ( -\frac{2}{5},\ -\frac{1}{5} ) $
D) $ ( \frac{2}{5},\ -\frac{1}{5} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
Ellipse is $ \frac{x^{2}}{\frac{1}{4}}+\frac{y^{2}}{\frac{1}{9}}=1 $
therefore $ a^{2}=\frac{1}{4} $ , $ b^{2}=\frac{1}{9} $
The equation of its tangent is $ 4xx’+9yy’=1 $
$ m=-\frac{4x’}{9y’}=\frac{8}{9} $
therefore $ x’=-2y’ $
and $ 4x{{’}^{2}}+9y{{’}^{2}}=1 $
therefore $ 4x{{’}^{2}}+9\frac{x{{’}^{2}}}{4}=1 $
therefore $ x’=\pm \frac{2}{5} $
When $ x’=\frac{2}{5},y’=-\frac{1}{5} $ and when $ x’=-\frac{2}{5},y’=\frac{1}{5} $
Hence points are $ ( \frac{2}{5},-\frac{1}{5} ) $ and $ ( -\frac{2}{5},\frac{1}{5} ) $ .
BETA
