Question

Let $5x^2+7y^2=140$, then $(\sqrt{14},\sqrt{10})$ is:

• A. Outside the ellipse
• B. Inside the ellipse
• C. On the ellipse
• D. Centre of the ellipse

Answer 1

The correct option is C On the ellipse

${5x}^2+{7y}^2=140$ (equation of ellipse)

putting $(\sqrt{14},\sqrt{10})$ in equation

$=5\times {\left(\sqrt{14}\right)}^2+7\times {\left(\sqrt{10}\right)}^2-140$

$=70+70-140$

$=0$

Hence, point lie on the ellipse.