If a double ordinate of the parabola y^2 = 4a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the parabola is $y^2 = 4ax$. The vertex is at $O(0, 0)$.Let the double ordinate be at $x = h$. Then $y^2 = 4ah$,so $y = \pm 2\sqrt

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The equation of the parabola is $y^2 = 4ax$. The vertex is at $O(0, 0)$.Let the double ordinate be at $x = h$. Then $y^2 = 4ah$,so $y = \pm 2\sqrt{ah}$.The length of the double ordinate is $2 	imes 2\sqrt{ah} = 4\sqrt{ah}$.Given the length is $8a$,we have $4\sqrt{ah} = 8a$,which implies $\sqrt{ah} = 2a$,so $ah = 4a^2$,hence $h = 4a$.The ends of the double ordinate are $P(4a, 4a)$ and $Q(4a, -4a)$.The slope of $OP$ is $m_1 = \frac{4a - 0}{4a - 0} = 1$.The slope of $OQ$ is $m_2 = \frac{-4a - 0}{4a - 0} = -1$.Since $m_1 	imes m_2 = 1 	imes (-1) = -1$,the lines $OP$ and $OQ$ are perpendicular to each other.Therefore,the angle between them is $90^\circ$.

If a double ordinate of the parabola $y^2 = 4ax$ has a length of $8a$,then the angle between the lines joining the vertex of the parabola to the ends of this double ordinate is ............... $^\circ$.

Similar Questions

The equation of the parabola whose vertex is at $(2, -1)$ and focus at $(2, -3)$ is

The equation of the normal to the parabola $y^2 = 4ax$ at the point $\left( \frac{a}{m^2}, \frac{2a}{m} \right)$ is:

$A$ ray of light travelling along the line $x = 2$ strikes a parabolic mirror with the equation $y^2 - 2y - 4x + 5 = 0$ and gets reflected from its surface. Then,the equation of the reflected ray may be:

If the line $y = 2x + k$ is a tangent to the curve $x^2 = 4y$,then $k$ is equal to

If the tangent at a point $P$ on a parabola makes an angle $\alpha$ with its focal distance,then the angle between the tangent and the axis of the parabola is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module