Let (h, k) lie on the circle C : x² + y² = 4 and the point (2h + 1, 3k + 2) lie on an ellipse with e

Q: Why do we use parametric coordinates $(2 \cos \theta, 2 \sin \theta)$ instead of just solving for $h$ or $k$ using $h^2 + k^2 = 4$?

That is a fantastic question! While you *could* solve for $h = \pm \sqrt{4 - k^2}$, you would immediately run into the nightmare of square roots and potential sign errors. Parametric coordinates are the 'secret weapon' of coordinate geometry. They allow us to convert a geometric constraint into a simple trigonometric identity, which is almost always cleaner and less prone to algebraic traps.

Q: How do I know which denominator is $a^2$ and which is $b^2$ in the ellipse equation?

In the standard equation of an ellipse, $\frac{(x - h')^2}{A} + \frac{(y - k')^2}{B} = 1$, the value $a^2$ is *always* the larger denominator. It represents the square of the semi-major axis. The smaller denominator is $b^2$, the square of the semi-minor axis. Always check which number is bigger before calculating eccentricity!

Q: Wait, why does the ellipse equation look shifted? The standard form is usually $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

You are absolutely right about the standard form! However, the transformation $x = 2h + 1$ and $y = 3k + 2$ introduces a shift. This is a 'translated' ellipse. The center of this ellipse is not at the origin $(0,0)$, but at $(1,2)$. The shape remains the same, just moved in the Cartesian plane.

Question

Let $(h, k)$ lie on the circle $C : x^2 + y^2 = 4$ and the point $(2h + 1, 3k + 2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac{5}{e^2}$ is equal to .........

Sigma Percentile · Accepted Answer

### The Geometry of Transformation: A Journey into the Locus

Welcome, fellow traveler on the path to JEE mastery. Today, we aren't just solving a problem; we are uncovering a hidden geometry.

We are going to take a simple circle and, through the power of algebraic transformation, watch it morph into an ellipse. This is the beauty of coordinate geometry—it is not just about numbers; it is about the dance of shapes.

### Phase 1: The Circle and the Parametric Bridge

We begin with the equation $x^2 + y^2 = 4$. This is our foundation, representing a circle centered at the origin with a radius of $r = 2$.

We are told that a point $(h, k)$ lies on this circle. To avoid the complexity of square roots, we use the 'Parametric Bridge'.

Any point on a circle of radius $r$ can be elegantly described by an angle $	heta$. We define:
$$h = 2 \cos 	heta$$
$$k = 2 \sin 	heta$$

This transforms the constraint $h^2 + k^2 = 4$ into the fundamental identity $\cos^2 	heta + \sin^2 	heta = 1$. It is the ultimate simplification.

### Phase 2: The Transformation Mapping

The problem introduces a new point $(x, y)$ defined by the transformation $x = 2h + 1$ and $y = 3k + 2$. Imagine this as a mapping where every point on our circle is stretched and shifted to a new location.

To find the locus of this new point, we express our parameter $	heta$ in terms of the new coordinates $x$ and $y$. Substituting our parametric definitions into the transformation equations, we obtain:
$$x = 2(2 \cos 	heta) + 1 = 4 \cos 	heta + 1$$
$$y = 3(2 \sin 	heta) + 2 = 6 \sin 	heta + 2$$

### Phase 3: The Elimination of the Parameter

Now, we isolate the trigonometric terms to prepare for the elimination of $	heta$:
$$\cos 	heta = \frac{x - 1}{4}$$
$$\sin 	heta = \frac{y - 2}{6}$$

Invoking the Pythagorean identity, $\cos^2 	heta + \sin^2 	heta = 1$, we substitute our expressions to arrive at the equation of the locus:
$$\left( \frac{x - 1}{4} ight)^2 + \left( \frac{y - 2}{6} ight)^2 = 1$$

This simplifies to the standard form:
$$\frac{(x - 1)^2}{16} + \frac{(y - 2)^2}{36} = 1$$

### Phase 4: The Ellipse and the Final Calculation

This equation represents an ellipse shifted to the center $(1, 2)$. Because $36 > 16$, the major axis is vertical, giving us $a^2 = 36$ and $b^2 = 16$.

The eccentricity $e$ is defined by the relation $e^2 = 1 - \frac{b^2}{a^2}$. Plugging in our values:
$$e^2 = 1 - \frac{16}{36} = 1 - \frac{4}{9} = \frac{5}{9}$$

Finally, the problem asks for the value of $\frac{5}{e^2}$. Substituting our result:
$$\frac{5}{5/9} = 5 	imes \frac{9}{5} = 9$$

The calculation is complete, and the elegance of the result is undeniable. You have successfully navigated the transformation; keep this intuition—that geometry is just algebra in disguise—and you will conquer any problem the JEE throws your way. The final answer is **9**.

Animated Solution for Mathematics - Conic Sections: Let $(h, k)$ lie on the circle $C : x^{2} + y^{2} = 4$ and the point $(2 h + 1, 3 k + 2)$ lie on an ellipse with eccentricity $e$ . Then the value of $\frac{5}{e ^{2}}$ is equal to .........

Enter Numerical Value:

Visualized Solution

The Given Circle $C$

Parametric Coordinates of $(h, k)$

Defining the New Point $(x, y)$

Substituting Parametric Values

Isolating $cos θ$ and $sin θ$

Applying the Identity $cos^{2} θ + sin^{2} θ = 1$

Formulating the Ellipse Equation

Identifying $a^{2}$ and $b^{2}$

Calculating Eccentricity $e^{2}$

Final Calculation of $\frac{5}{e ^{2}}$

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

The Geometry of Transformation

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

The Given Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Parametric Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (h,k)

Defining the New Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x,y)