Consider the family of circles x² + y² = r², 2

Q: Wait, why is the circle condition redundant? Does it not restrict the tangent?

That is a brilliant observation! In many problems, extra information is provided to test your ability to filter noise. For the ellipse $\frac{x^2}{25} + \frac{y^2}{4} = 1$, the perpendicular distance $p$ of a tangent from the origin is given by $p = \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta}$. Since $b < p < a$, we have $2 < p < 5$. Because the family of circles covers all radii $r$ in $(2, 5)$, every single tangent to the ellipse is guaranteed to be a tangent to one of the circles. The circle condition is satisfied by default for every tangent!

Q: Why use the parametric form $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$ instead of $y = mx + c$?

You could use $y = mx + c$, but it often leads to messy quadratic equations in $m$ and requires handling the slope $m$ carefully. The parametric form is the 'golden key' for locus problems involving ellipses. It naturally introduces a single parameter $\theta$, which is easily eliminated using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$. It turns a complex algebraic struggle into a clean, elegant derivation.

Q: How do I know if the locus is valid for the entire curve?

Great question! The problem specifies the tangent is in the first quadrant. This implies that the intercepts $A$ and $B$ must be positive, which restricts $\theta$ to the interval $(0, \pi/2)$. Consequently, the locus is only a portion of the resulting curve. However, in JEE problems, unless explicitly asked for the domain, providing the general equation of the locus is the standard expectation.

Question

Consider the family of circles $x^2 + y^2 = r^2, 2 < r < 5$. If in the first quadrant, the common tangent to a circle of this family and the ellipse $4x^2 + 25y^2 = 100$ meets the co-ordinate axes at $A$ and $B$, then find the equation of the locus of the mid-point of $AB$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the edge of a vast coordinate plane, looking at an ellipse defined by $4x^2 + 25y^2 = 100$. You are told that a family of circles $x^2 + y^2 = r^2$ exists, with $r$ dancing between $2$ and $5$.

At first glance, this feels like a complex interaction between two distinct geometric entities. But here is the secret: the circle condition is a phantom. It is a beautiful, distracting veil.

As we dive into the math, we will see that the ellipse itself dictates the behavior of these tangents, and the circles are merely silent observers. Let us begin by standardizing our ellipse.

By dividing the equation $4x^2 + 25y^2 = 100$ by $100$, we transform it into the elegant standard form:

$$ \frac{x^2}{25} + \frac{y^2}{4} = 1 $$

Here, the semi-major axis $a$ is $5$, and the semi-minor axis $b$ is $2$. This is our foundation.

### The Parametric Compass

When dealing with tangents to an ellipse, the slope-intercept form $y = mx \pm \sqrt{a^2m^2 + b^2}$ can be cumbersome. Instead, we embrace the parametric form.

Any point on this ellipse can be represented as $(5 \cos 	heta, 2 \sin 	heta)$. The tangent at this point is given by the beautiful equation:

$$ \frac{x \cos 	heta}{5} + \frac{y \sin 	heta}{2} = 1 $$

This equation is our compass. It tells us exactly how the tangent line behaves as $	heta$ varies.

To find the intercepts $A$ and $B$, we simply set the variables to zero. For point $A$ on the x-axis, we set $y = 0$, yielding $x = \frac{5}{\cos 	heta}$. Thus, $A \equiv (\frac{5}{\cos 	heta}, 0)$.

For point $B$ on the y-axis, we set $x = 0$, yielding $y = \frac{2}{\sin 	heta}$. Thus, $B \equiv (0, \frac{2}{\sin 	heta})$. We have captured the endpoints of our tangent segment.

### The Midpoint Bridge

Now, we seek the locus of the midpoint $M(h, k)$ of the segment $AB$. The midpoint formula is our bridge:

$$ h = \frac{5}{2 \cos 	heta}, \quad k = \frac{1}{\sin 	heta} $$

We are now at the threshold of the final solution. We have $h$ and $k$ in terms of $	heta$, but the locus requires an equation in $x$ and $y$ alone. We must eliminate $	heta$.

Rearranging our equations, we find $\cos 	heta = \frac{5}{2h}$ and $\sin 	heta = \frac{1}{k}$. The final step is the most satisfying part of the journey.

### Final Calculation

We invoke the fundamental identity of trigonometry: $\cos^2 	heta + \sin^2 	heta = 1$. Substituting our expressions, we get:

$$ \left( \frac{5}{2h} ight)^2 + \left( \frac{1}{k} ight)^2 = 1 $$

Expanding this, we arrive at:

$$ \frac{25}{4h^2} + \frac{1}{k^2} = 1 $$

Replacing $(h, k)$ with $(x, y)$ and multiplying by $4$ to clean up the fractions, we reach our destination:

$$ \frac{25}{x^2} + \frac{4}{y^2} = 4 $$

This is the equation of the locus. It is not just a collection of symbols; it is the mathematical signature of the midpoint's path. You have successfully navigated the trap, utilized the parametric power, and arrived at the **elegant truth**.

Visualized Solution

Visualizing the Setup

The JEE Trap Revealed

Standardizing the Ellipse

The Parametric Tangent

Finding the X-Intercept (Point A)

Finding the Y-Intercept (Point B)

Setting up the Midpoint M

Calculating Coordinates of M

Isolating the Parameter $θ$

Eliminating $θ$ using Identity

Expanding the Equation

The Final Locus Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Compass

The Midpoint Bridge

Final Calculation

Similar Questions

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If the tangents drawn to the hyperbola $4 y^{2} = x^{2} + 1$ intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

Two circles each of radius 5 units touch each other at the point $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , and $C_{1} (α, β)$ and $C_{2} (γ, δ)$ , $C_{1} \neq = C_{2}$ are their centres, then $∣ (α + β) (γ + δ) ∣$ is equal to .

Two circles, each of radius 5 units, touch each other at $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , find the equation of the circles.

If a tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the tangents at the extremities of its major axis at $B$ and $C$ , then the circle with $B C$ as diameter passes through the point:

From the origin chords are drawn to the circle $(x - 1)^{2} + y^{2} = 1$ . The equation of the locus of the mid-points of these chords is .........

Find the equation of the common tangent in 1st quadrant to the circle $x^{2} + y^{2} = 16$ and the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1$ . Also find the length of the intercept of the tangent between the coordinate axes.