If the line $3x-2y+12=0$ intersects the parabola $4y=3x^2$ at the points $A$ and $B$, then at the vertex of the parabola, the line segment $AB$ subtends an angle equal to

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Substituting $y=\frac{3x+12}2$ into $4y=3x^2$: $x^2-2x-8=0$, so $A(4,12)$ and $B(-2,3)$. The vertex is $O(0,0)$: $\vec{OA}=(4,12)$, $\vec{OB}=(-2,3)$. $	an	heta=\frac{|4\cdot3-12\cdot(-2)|}{4(-2)+12\cdot3}=\frac{36}{28}=\frac97$. Answer (2).

Q. If the line $3x-2y+12=0$ intersects the parabola $4y=3x^2$ at the points $A$ and $B$, then at the vertex of the parabola, the line segment $AB$ subtends an angle equal to

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?