- Home
- Standard 11
- Mathematics
On the ellipse $4{x^2} + 9{y^2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are
$\left( {\frac{2}{5},\;\frac{1}{5}} \right)$
$\left( { - \frac{2}{5},\;\frac{1}{5}} \right)$
$\left( { \frac{2}{5},\; - \frac{1}{5}} \right)$
$(b)$ and $(c)$ both
Solution
(d) Ellipse is $\frac{{{x^2}}}{{\frac{1}{4}}} + \frac{{{y^2}}}{{\frac{1}{9}}} = 1$
==> ${a^2} = \frac{1}{4}$, ${b^2} = \frac{1}{9}$
The equation of its tangent is $4xx' + 9yy' = 1$
$m = – \frac{{4x'}}{{9y'}} = \frac{8}{9}$
==> $x' = – 2y'$
and $4x{'^2} + 9y{'^2} = 1$
==> $4x{'^2} + 9\frac{{x{'^2}}}{4} = 1$
==> $x' = \pm \frac{2}{5}$
When $x' = \frac{2}{5},\,\,y' = – \frac{1}{5}$
and when $x' = – \frac{2}{5},\,\,\,y' = \frac{1}{5}$
Hence points are $\left( {\frac{2}{5},\,\, – \frac{1}{5}} \right)\,$
and $\left( { – \frac{2}{5},\,\,\frac{1}{5}} \right)$.
