1. Tardigrade
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  3. Mathematics
  4. Consider an ellipse E: ( x 2/16)+( y 2/12)=1 and a parabola P whose vertex is (-√3, 0) and focus is the origin. The area of the triangle formed by the normals to the ellipse and the parabola at their point of intersection and the x-axis is

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Q. Consider an ellipse $E : \frac{ x ^2}{16}+\frac{ y ^2}{12}=1$ and a parabola $P$ whose vertex is $(-\sqrt{3}, 0)$ and focus is the origin.
The area of the triangle formed by the normals to the ellipse and the parabola at their point of intersection and the $x$-axis is

Conic Sections

Solution:

Equation of parabola is $y^2=4 \sqrt{3}(x+\sqrt{3})$
Normal to the parabola at $B$ meets the $x$ axis at $N (2 \sqrt{3}, 0)$
and normal to the ellipse at B is the y-axis.
$\therefore $ Required area $=\frac{1}{2} \times 2 \sqrt{3} \times 2 \sqrt{3}=6$ sq-units

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