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

Question

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 (A) 2 sq. units (B) 4 sq. units (C) 6sq. units (D) 8 √3 sq units.

Tardigrade Admin · Accepted Answer

Equation of parabola is y2=4 √3(x+√3) Normal to the parabola at B meets the x axis at N (2 √3, 0) and normal to the ellipse at B is the y-axis. ∴ Required area =(1/2) × 2 √3 × 2 √3=6 sq-units

Q. Consider an ellipse $E : \frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1$ 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

2 sq. units

4 sq. units

6sq. units

$83$ sq units.

Equation of parabola is $y^{2} = 43 (x + 3)$
Normal to the parabola at $B$ meets the $x$ axis at $N (23, 0)$
and normal to the ellipse at B is the y-axis.
$∴$ Required area $= \frac{1}{2} \times 23 \times 23 = 6$ sq-units

Q. Consider an ellipse E:16x2​+12y2​=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