If two normals are drawn from a point to the | vedclass

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

Question

(A) Let the point be $(h, k)$. The equation of a normal to the parabola $y^{2} = 4ax$ is given by $y = mx - 2am - am^{3}$.Since it passes through $(h,

Vedclass · Accepted Answer

$y^{2} = 4ax$

Vedclass · Answer

(A) Let the point be $(h, k)$. The equation of a normal to the parabola $y^{2} = 4ax$ is given by $y = mx - 2am - am^{3}$.Since it passes through $(h, k)$,we have $k = mh - 2am - am^{3}$,which simplifies to $am^{3} + m(2a - h) + k = 0$.Let the slopes of the normals be $m_1, m_2, m_3$. The product of the roots is $m_1 m_2 m_3 = -k/a$.Given that the normals make angles $\alpha$ and $\beta$ with the axis,their slopes are $m_1 = 	an \alpha$ and $m_2 = 	an \beta$. We are given $m_1 m_2 = 2$.Substituting this into the product of roots,we get $2 m_3 = -k/a$,so $m_3 = -k/(2a)$.Since $m_3$ is a root of the cubic equation $am^{3} + m(2a - h) + k = 0$,we substitute $m = -k/(2a)$:$a(-k/(2a))^{3} + (-k/(2a))(2a - h) + k = 0$$-k^{3}/(8a^{2}) - k + kh/(2a) + k = 0$$-k^{3}/(8a^{2}) + kh/(2a) = 0$$k^{2} = 4ah$.Replacing $(h, k)$ with $(x, y)$,the locus is $y^{2} = 4ax$.

If two normals are drawn from a point to the parabola $y^{2} = 4ax$ such that they make angles $\alpha$ and $\beta$ with the axis,and $\tan \alpha \cdot \tan \beta = 2$,find the locus of the point.

Similar Questions

The length of the latus rectum of the parabola $(x-2)^2+(y-3)^2=\frac{1}{25}(3x-4y+7)^2$ is

Let chord $PQ$ of length $3\sqrt{13}$ of the parabola $y^2 = 12x$ be such that the ordinates of points $P$ and $Q$ are in the ratio $1:2$. If the chord $PQ$ subtends an angle $\alpha$ at the focus of the parabola, then $\sin \alpha$ is equal to:

$A$ particle is moving in the $xy$-plane along a curve $C$ passing through the point $(3, 3)$. The tangent to the curve $C$ at the point $P$ meets the $x$-axis at $Q$. If the $y$-axis bisects the segment $PQ$,then $C$ is a parabola with

The Cartesian equation of the parabola $x = -2 + 2t^2$,$y = 2 + 4t$ is

If all the vertices of an equilateral triangle lie on the parabola $y^2=16x$ and one of them coincides with the vertex of that parabola,then the length of the side of that triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module