Question

On the ellipse $4x^2+9y^2=1$, the point at which the tangents are parallel to the line $8x=9$ are

• A. $(\frac{2}{5},\frac{1}{5})$
• B. $(-\frac{2}{5},\frac{1}{5})$
• C. $(-\frac{2}{5},-\frac{1}{5})$
• D. $(\frac{2}{5},-\frac{1}{5})$

Answer 1

Answer: (B), (D)

The correct options are
B $(-\frac{2}{5},\frac{1}{5})$
D $(\frac{2}{5},-\frac{1}{5})$

Ellipse is $\frac{x^2}{\frac{1}{4}}+\frac{y^2}{\frac{1}{9}}=1\Rightarrow a^2=\frac{1}{4},b^2=\frac{1}{9}$

The equation of its tangent is $4x{x}^{\prime }+9y{y}^{\prime }=1$

$\therefore m=-\frac{4x^{\prime }}{9y^{\prime }}=\frac{8}{9}\Rightarrow x^{\prime }=-2y^{\prime }$

and $4{x^{\prime }}^2+9{y^{\prime }}^2=1\Rightarrow 4{x^{\prime }}^2+\frac{9{x^{\prime }}^2}{4}=1\Rightarrow x^{\prime }=\pm \frac{2}{5}$

when $x^{\prime }=\frac{2}{5},y^{\prime }=-\frac{1}{5}$ and when $x^{\prime }=-\frac{2}{5},y^{\prime }=\frac{1}{5}$

Hence points are $(\frac{2}{5},-\frac{1}{5})$ and $(-\frac{2}{5},\frac{1}{5})$.