Let O be the vertex of the parabola x^{2}=4y | vedclass

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

Question

(D) Let $Q = (2t, t^2)$ be a point on the parabola $x^2 = 4y$. The vertex $O$ is $(0, 0)$.The point $P(h, k)$ divides $OQ$ in the ratio $2:3$. Using t

Vedclass · Accepted Answer

$5x - 4y + 3 = 0$

Vedclass · Answer

(D) Let $Q = (2t, t^2)$ be a point on the parabola $x^2 = 4y$. The vertex $O$ is $(0, 0)$.The point $P(h, k)$ divides $OQ$ in the ratio $2:3$. Using the section formula:$h = \frac{2(2t) + 3(0)}{2+3} = \frac{4t}{5} \Rightarrow t = \frac{5h}{4}$$k = \frac{2(t^2) + 3(0)}{2+3} = \frac{2t^2}{5} = \frac{2}{5} \left(\frac{5h}{4}\right)^2 = \frac{2}{5} \cdot \frac{25h^2}{16} = \frac{5h^2}{8}$Thus, the locus $C$ is $8k = 5h^2$, or $5x^2 = 8y$.The equation of the chord of the parabola $5x^2 = 8y$ bisected at $(x_1, y_1) = (1, 2)$ is given by $T = S_1$, where $T = 5x(x_1) - 4(y + y_1)$ and $S_1 = 5x_1^2 - 8y_1$.Substituting the values:$5x(1) - 4(y + 2) = 5(1)^2 - 8(2)$$5x - 4y - 8 = 5 - 16$$5x - 4y - 8 = -11$$5x - 4y + 3 = 0$

Let $O$ be the vertex of the parabola $x^{2}=4y$ and $Q$ be any point on it. Let the locus of the point $P$, which divides the line segment $OQ$ internally in the ratio $2:3$, be the conic $C$. Then the equation of the chord of $C$, which is bisected at the point $(1, 2)$, is:

Similar Questions

The number of normals drawn to the parabola $y^2=4x$ from the point $(1,0)$ is

The equation of the tangent at $(-4, -4)$ on the curve $x^2 = -4y$ is

Variable straight lines $y=mx+c$ make intercepts on the curve $y^2-4ax=0$ which subtend a right angle at the origin. Then the point of concurrence of these lines $y=mx+c$ is

The length of the latus rectum of the parabola ${y^2} = 5x + 4y + 1$ is

If $\overrightarrow{AB}$ is the focal chord of the parabola $y^2=16x$ and $A=(1,-4)$,then the equation of the normal to the parabola at the point $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module