Problem 60 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The tangent at the point ' \(\alpha\) ' on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) meets the auxiliary circle in two points which subtend a right angle at the centre. The eccentricity of the ellipse is (A) \(\frac{1}{\sqrt{1+\sin ^{2} \alpha}}\) (B) \(\frac{1}{\sqrt{1+\cos ^{2} \alpha}}\) (C) \(\sqrt{1+\sin ^{2} \alpha}\) (D) none of these

Step-by-Step Solution

Answer

The eccentricity is \(\frac{1}{\sqrt{1+\sin^2 \alpha}}\).

Step 1: Understand the problem

We need to find the eccentricity of the ellipse given the condition that the tangent at the point ' \(\alpha\) ' on the ellipse intersects the auxiliary circle and forms a right angle at the center.

Step 2: Review ellipse and circle equations

The given ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), and its auxiliary circle is \(x^2 + y^2 = a^2\). The tangent of the ellipse at point \((x_1, y_1)\) is given by \(\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1\).

Step 3: Find equation of tangent at point \(\alpha\)

Assume the point on the ellipse at angle \(\alpha\) is \((a\cos \alpha, b\sin \alpha)\). The tangent equation is \(\frac{x \cdot a \cos \alpha}{a^2} + \frac{y \cdot b \sin \alpha}{b^2} = 1\), simplifying to \(x \cos \alpha / a + y \sin \alpha / b = 1\).

Step 4: Analyze the intersection with the auxiliary circle

The auxiliary circle's equation is \(x^2 + y^2 = a^2\). Substituting the tangent's equation into this circle's equation, solve for \(x\) and \(y\). Substitute \(x = a \cos(\alpha + \theta)\) and \(y = a \sin(\alpha + \theta)\).

Step 5: Apply the condition for orthogonality

Two points \((x_1, y_1)\) and \((x_2, y_2)\) subtend a right angle at the center if their dot product is zero: \(x_1 x_2 + y_1 y_2 = 0\). Solve this equation using step 4's substitutions.

Step 6: Derive the expression for eccentricity

Using the orthogonality condition, simplify to find eccentricity \(e = \sqrt{1 - \frac{b^2}{a^2}}\). Use the condition found in step 5 to deduce \(e = \frac{1}{\sqrt{1+\sin^2 \alpha}}\).

Step 7: Verify the eccentricity

Check steps and ensure calculations meet the problem's requirements, confirming the value for \(e\). This value matches option (A).

Key Concepts

Eccentricity of Ellipse

The eccentricity of an ellipse is a measure of how much the ellipse deviates from being a perfect circle. It is a number between 0 and 1, and you can think of it as a scale of roundness:

• An eccentricity of 0 represents a perfect circle.
• An eccentricity close to 1 indicates that the ellipse is more stretched out.

For any given ellipse described by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the eccentricity \(e\) can be calculated as:\[e = \sqrt{1 - \frac{b^2}{a^2}}.\]This formula arises from the relationship between the semi-major axis \(a\) and the semi-minor axis \(b\) of the ellipse. The semi-major axis is the longest radius, and the semi-minor axis is the shortest. The eccentricity gives a sense of the shape, determining how elongated the ellipse is. In this exercise, we use additional problem conditions to further simplify and determine the eccentricity in terms of angle \(\alpha\).

Auxiliary Circle

An auxiliary circle of an ellipse is an important geometric tool that assists in analyzing properties such as tangency and orthogonality. This circle is centered at the origin and has a radius equal to the semi-major axis of the ellipse, \(a\).
The equation for the auxiliary circle is:\[x^2 + y^2 = a^2.\]This circle helps in bridging the properties of ellipses and circles, particularly in understanding how tangents behave. In this problem, the auxiliary circle is used because the tangent to the ellipse, when extended, intersects this circle. This allows one to leverage circular geometry properties to deduce information about the ellipse, such as where tangents meet or how they are angled in relation to the center.

Orthogonality Condition

The orthogonality condition is employed when two points on the auxiliary circle subtend a right angle at the center. For any two points \((x_1, y_1)\) and \((x_2, y_2)\) on a circle, if they subtend a right angle at the circle's center, their dot product equals zero:
\[x_1 x_2 + y_1 y_2 = 0.\]This is a result of the property that perpendicular vectors have a dot product of zero. This condition is particularly useful in determining the specific relationship between points on an ellipse and its auxiliary circle that will satisfy given geometric conditions. By solving these equations, it helps in isolating factors like angles and distances within the problem, ultimately aiding in finding the eccentricity of the ellipse.

Tangent Equation

The tangent to an ellipse at a given point represents a line that just touches the ellipse at that exact point without intersecting it. For the ellipse given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the tangent line at a point \((a\cos \alpha, b\sin \alpha)\) is derived as:\[x \cos \alpha / a + y \sin \alpha / b = 1.\]This equation ensures that for all points \((x, y)\) on this line, the relationship holds true. It serves as a foundational basis for solving further parts of the problem by relating the ellipse's geometric properties to other bodies like the auxiliary circle. The tangent's behavior, particularly where and how it intersects other figures like circles, often leads to deeper insights into the ellipse's intrinsic attributes, such as its eccentricity.