Let O be the vertex of the parabola x^2=4 y and Q be any point on it.

Question

Let O be the vertex of the parabola $x^2=4 y$ 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 bisecte...

Accepted Answer

Let the vertex of the parabola ${x}^{2}=4y$ be the origin $O(0,0)$.

Let $Q$ be any point on the parabola. We can parameterize the coordinates of $Q$ as $(2t,{t}^{2})$ for some real parameter $t$. Satisfying ${x}^{2}=(2t{)}^{2}=4{t}^{2}$ and $4y=4({t}^{2})$.

Let $P(x,y)$ be the point that divides the line segment $OQ$ internally in the ratio $2:3$.

Using the section formula, the coordinates of $P$ are given by:

$x=\frac{2\cdot {x}_{Q}+3\cdot {x}_{O}}{2+3}=\frac{2(2t)+3(0)}{5}=\frac{4t}{5}$

$y=\frac{2\cdot {y}_{Q}+3\cdot {y}_{O}}{2+3}=\frac{2({t}^{2})+3(0)}{5}=\frac{2{t}^{2}}{5}$

To find the locus of $P$ (the conic $C$), we eliminate the parameter $t$.

From the first equation, we have $t=\frac{5x}{4}$.

Substituting this into the second equation:

$y=\frac{2}{5}{(\frac{5x}{4})}^{2}$

$y=\frac{2}{5}\cdot \frac{25{x}^{2}}{16}$

$y=\frac{5{x}^{2}}{8}$

$5{x}^{2}=8y$

So, the equation of the conic $C$ is $5{x}^{2}-8y=0$. This is a parabola.

We need to find the equation of the chord of $C$ which is bisected at the point $({x}_{1},{y}_{1})=(1,2)$.

The equation of a chord of a conic $S=0$ bisected at a point $({x}_{1},{y}_{1})$ is given by $T={S}_{1}$.

Here, $S\equiv 5{x}^{2}-8y$.

${S}_{1}$ is the value of $S$ at $(1,2)$:

${S}_{1}=5(1{)}^{2}-8(2)=5-16=-11$

$T$ is the expression obtained by replacing ${x}^{2}$ with $x{x}_{1}$ and $y$ with $\frac{y+{y}_{1}}{2}$:

$T=5x(1)-8(\frac{y+2}{2})$

$T=5x-4(y+2)$

$T=5x-4y-8$

Equating $T={S}_{1}$:

$5x-4y-8=-11$

$5x-4y+3=0$

The equation of the chord is $5x-4y+3=0$.

Option A

$5x-4y+3=0$

Solutions · 1