Let O be the vertex of the parabola x² = 4y and Q be any point on it. Let the locus of the point P,

Q: Wait, why do we use $T = S_1$ for the chord? Can't we just find the slope and use $y - y_1 = m(x - x_1)$?

You absolutely can! However, finding the slope $m$ requires you to differentiate the conic or use the property that the slope of the chord bisected at $(x_1, y_1)$ is $-x_1 / (2a)$ for $x^2 = 4ay$. The $T = S_1$ method is a powerful, direct, and elegant shortcut that saves precious time in the JEE Advanced exam. It is a standard tool you should master.

Q: Why is the locus of $P$ also a parabola? Is there a geometric reason?

Yes! The transformation we applied is a linear scaling (a dilation). When you take a parabola and scale its coordinates by constant factors, the resulting curve remains a parabola. The geometric structure is preserved; only the 'width' or 'steepness' of the parabola changes.

Q: I got $x^2 = \frac{8}{5}y$. How do I know if this is the correct conic $C$?

Great job! You can verify this by taking a point on the original parabola, say $(2, 1)$, and finding the corresponding point $P$ on the new locus. If your derived equation $x^2 = \frac{8}{5}y$ holds true for the transformed coordinates, you have successfully found the correct locus.

Question

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 :

Options:

(A) $5x - 4y + 3 = 0$

(B) $5x - y - 3 = 0$

(C) $4x - 5y + 6 = 0$

(D) $x - 2y + 3 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking at the graceful curve of the parabola $x^2 = 4y$. It is a classic, a foundational shape in our study of conic sections.

To begin, we parameterize our point $Q$ on the parabola. Instead of dealing with messy $(x, y)$ coordinates, we use the parameter $t$. By setting $Q = (2t, t^2)$, we satisfy the equation $x^2 = 4y$ because $(2t)^2 = 4(t^2)$.

### The Dance of the Section Formula

Consider the line segment connecting the origin $O(0, 0)$ and our point $Q(2t, t^2)$. We are told that a point $P(h, k)$ divides this segment in a $2:3$ ratio.

Using the section formula, we find the coordinates of $P$:

$$h = \frac{2(2t) + 3(0)}{2+3} = \frac{4t}{5}$$

$$k = \frac{2(t^2) + 3(0)}{2+3} = \frac{2t^2}{5}$$

To find the locus, we must eliminate $t$. From $h = \frac{4t}{5}$, we get $t = \frac{5h}{4}$. Substituting this into our expression for $k$:

$$k = \frac{2}{5} \left( \frac{5h}{4} \right)^2 = \frac{2}{5} \cdot \frac{25h^2}{16} = \frac{5h^2}{8}$$

Replacing $h$ and $k$ with $x$ and $y$, we arrive at the equation of our new conic $C$:

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

### The Final Chord

The problem now asks for the equation of a chord of this conic that is bisected at the point $M(1, 2)$. This is a classic scenario where the $T = S_1$ formula shines.

For a conic $S = 0$, the equation of the chord bisected at $(x_1, y_1)$ is $T = S_1$, where $T$ is the tangent expression and $S_1$ is the value of the conic at that point. For our parabola $x^2 - \frac{8}{5}y = 0$, the tangent expression $T$ at $(1, 2)$ is:

$$T = x(1) - \frac{4}{5}(y + 2)$$

The value $S_1$ is calculated as:

$$S_1 = (1)^2 - \frac{8}{5}(2) = 1 - \frac{16}{5} = -\frac{11}{5}$$

Equating $T = S_1$, we get:

$$x - \frac{4}{5}(y + 2) = -\frac{11}{5}$$

Multiplying by $5$ to clear the denominators, we have $5x - 4(y + 2) = -11$. This simplifies to $5x - 4y - 8 = -11$, which yields the final result:

**$5x - 4y + 3 = 0$**

Select Answer:

Visualized Solution

Visualizing the Parabola $x^{2} = 4 y$

Parametric Form of $Q$

Defining Point $P$ using Section Formula

Coordinates of $P$ in terms of $t$

Eliminating Parameter $t$

Finding the Locus of $P$ (Conic $C$ )

Visualizing the New Conic $C$

The $T = S_{1}$ Concept

Applying $T = S_{1}$ to Conic $C$

Equating $T$ and $S_{1}$

Simplifying the Equation

Final Equation of the Chord

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Dance of the Section Formula

The Final Chord

Similar Questions

Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2} = 4 x$ be the curve $S$ . Let $P$ be any point on $S$ . Then the locus of the point, which internally divides $O P$ in the ratio $3 : 1$ , is :

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If $a \neq = 0$ and the line $2 b x + 3 cy + 4 d = 0$ passes through the points of intersection of the parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ , then

Select Answer:

Visualized Solution

Visualizing the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2=4y

Parametric Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Defining Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P using Section Formula

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Eliminating Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Finding the Locus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (Conic .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C)

Visualizing the New Conic .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=S1​ Concept

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=S1​ to Conic .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​