Question

The point of intersection of the two tangent to the ellipse $2x^2+3y^2=6$ at the ends of latusrectum is:

• A. $(3,0)$
• B. $(\frac{7}{2},0)$
• C. $(\frac{9}{2},0)$
• D. $(4,0)$

Answer 1

Answer: (A)

$(3,0)$

Let coordinate of two ends of latus rectum be $(x_1,y_1)$ and $(x_2,y_2).$

$e=$ eccentricity $=\sqrt{1-{\left(\frac{b}{a}\right)}^2}$

$\Rightarrow e=\sqrt{\frac{1}{3}}$

$x_1=x_2=ae=1$

Putting $\left(1\right)$ into the equation of ellipse we will get value of $y_1$ and $y_2$

$y_1=\frac{2}{\sqrt{3}}$

$y_2=\frac{-2}{\sqrt{3}}$

Equation of two tangents are

$2x{x}_1+3y{y}_1=6\Rightarrow 2x+2\sqrt{3}y=6$

$2x{x}_2+3y{y}_2=6\Rightarrow 2x-2\sqrt{3}y=6$

Solving both equations we get,

$x=3$ and $y=0$