If tangents are drawn to the ellipse x² + 2y² = 2 at all points on the ellipse other than its four v

Q: Wait, why do we exclude the vertices of the ellipse?

Great observation! At the vertices (like $(\sqrt{2}, 0)$ or $(0, 1)$), the tangent is either vertical or horizontal. A vertical line is parallel to the y-axis and never intersects it, meaning the intercept is at infinity. Since we cannot define a finite midpoint for a segment with an infinite intercept, we must exclude these four points.

Q: Why can't we just use the slope-intercept form $y = mx \pm \sqrt{a^2m^2 + b^2}$?

You absolutely can! However, in JEE Advanced, time is your most precious resource. Using the slope-intercept form often leads to messy quadratic equations in $m$ and requires more algebraic manipulation. The parametric form $P(a \cos \theta, b \sin \theta)$ is the 'elegant' path because it naturally leads to the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$, which simplifies the locus calculation significantly.

Q: How do I know if my final equation is correct?

The final locus equation should be in terms of $x$ and $y$. Once you have the relation $\frac{1}{2h^2} + \frac{1}{4k^2} = 1$, simply replace $h$ with $x$ and $k$ with $y$. If your result matches one of the given options, you have successfully traced the path of the midpoint. Always double-check your arithmetic when squaring the terms!

Question

If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ 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 :

Options:

(A) $\frac{x^2}{2} + \frac{y^2}{4} = 1$

(B) $\frac{x^2}{4} + \frac{y^2}{2} = 1$

(C) $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$

(D) $\frac{1}{4x^2} + \frac{1}{2y^2} = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given ellipse is defined by the equation $x^2 + 2y^2 = 2$. To bring this into standard form, we divide the entire equation by $2$:

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

Here, the semi-major axis is $a = \sqrt{2}$ and the semi-minor axis is $b = 1$. This ellipse serves as the boundary for our sliding tangent line.

### The Power of Parameterization

To simplify the geometry, we represent any point $P$ on the ellipse using the parameter $	heta$:

$$P = (\sqrt{2} \cos 	heta, \sin 	heta)$$

As $	heta$ varies, $P$ traces the entire ellipse. We assume $	heta$ is such that the tangent is not parallel to the coordinate axes to ensure the intercepts remain finite.

### The Tangent Equation

The equation of a tangent to an ellipse at point $(x_1, y_1)$ is given by $\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$. Substituting our parametric coordinates, we obtain:

$$\frac{x(\sqrt{2} \cos 	heta)}{2} + \frac{y(\sin 	heta)}{1} = 1$$

Simplifying this expression, the equation of the sliding rod (the tangent line) becomes:

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

### The Intercepts and the Midpoint

To find the intercepts, we set one coordinate to zero. The x-intercept $A$ occurs at $y=0$, giving $x = \frac{\sqrt{2}}{\cos 	heta}$. The y-intercept $B$ occurs at $x=0$, giving $y = \frac{1}{\sin 	heta}$.

Let $M(h, k)$ be the midpoint of the segment $AB$. Using the midpoint formula, we find:

$$h = \frac{1}{\sqrt{2} \cos 	heta} \quad 	ext{and} \quad k = \frac{1}{2 \sin 	heta}$$

### Final Synthesis

To find the locus, we isolate the trigonometric functions: $\cos 	heta = \frac{1}{\sqrt{2} h}$ and $\sin 	heta = \frac{1}{2 k}$. We then apply the Pythagorean identity $\cos^2 	heta + \sin^2 	heta = 1$:

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

Expanding the squares, we arrive at the final equation for the locus:

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

Replacing $(h, k)$ with $(x, y)$, the path traced by the midpoint is:

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

Animated Solution for Mathematics - Conic Sections: 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 :

Select Answer:

Visualized Solution

Visualizing the Ellipse $x^{2} + 2 y^{2} = 2$

Parametric Point $P (2 cos θ, sin θ)$

Equation of the Tangent

Finding Intercepts $A$ and $B$

Defining the Midpoint $M (h, k)$

Isolating the Parameter $θ$

Eliminating $θ$ using $cos^{2} θ + sin^{2} θ = 1$

Squaring and Adding Equations

The Final Locus $\frac{1}{2 x ^{2}} + \frac{1}{4 y ^{2}} = 1$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of Parameterization

The Tangent Equation

The Intercepts and the Midpoint

Final Synthesis

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Select Answer:

Visualized Solution

Visualizing the Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+2y2=2

Parametric Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2​cosθ,sinθ)

Equation of the Tangent

Finding Intercepts .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Defining the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M(h,k)

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

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2θ+sin2θ=1