Let the line y = mx and the ellipse 2x² + y² = 1 intersect a point P in the first quadrant. If the n

Q: Wait, why can't we just use the standard line equation $y = mx + c$ and find the intersection?

That is a common trap! While you could try, you don't know the slope $m$ of the normal line initially. The normal is defined by the point of contact $P(x_1, y_1)$ on the ellipse. By using the point-slope form, we link the geometry of the ellipse directly to the line, which is much more efficient.

Q: Why do we reject the negative root when calculating $y_1$?

Great catch! The problem explicitly states that point $P$ lies in the first quadrant. In the first quadrant, both $x$ and $y$ coordinates must be positive. Always keep an eye on the constraints given in the problem statement—they are your best friends for eliminating extraneous solutions.

Q: Is there a faster way using parametric coordinates?

Absolutely! You could use $x = \frac{1}{\sqrt{2}}\cos\theta$ and $y = \sin\theta$. While it might look elegant, the algebraic method we used is often less prone to trigonometric identity errors during the exam. Both paths lead to the same truth, but choose the one where you feel most confident.

Question

Let the line $y = mx$ and the ellipse $2x^2 + y^2 = 1$ intersect a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(-\frac{1}{3\sqrt{2}}, 0)$ and $(0, \beta)$, then $\beta$ is equal to :

Options:

(A) $\frac{2}{\sqrt{3}}$

(B) $\frac{2}{3}$

(C) $\frac{2\sqrt{2}}{3}$

(D) $\frac{\sqrt{2}}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The ellipse is defined by the equation $2x^2 + y^2 = 1$. We are considering a point $P(x_1, y_1)$ located in the first quadrant, which serves as the origin point for a normal line.

Our objective is to determine the $y$-intercept, denoted as $\beta$, of this normal line.

### The Anatomy of the Normal

To derive the equation of the normal, we first determine the slope of the tangent at $P$. Differentiating the ellipse equation $2x^2 + y^2 = 1$ with respect to $x$ yields:

$$4x + 2y \frac{dy}{dx} = 0$$

Rearranging this expression, we find the slope of the tangent, $m_t$, at point $P(x_1, y_1)$:

$$m_t = -\frac{2x_1}{y_1}$$

Since the normal is perpendicular to the tangent, its slope $m_n$ is the negative reciprocal of $m_t$:

$$m_n = \frac{y_1}{2x_1}$$

### Building the Bridge

Using the point-slope form, the equation of the normal line is given by:

$$(y - y_1) = \frac{y_1}{2x_1}(x - x_1)$$

We are given that this line passes through the $x$-intercept $(-\frac{1}{3\sqrt{2}}, 0)$. Substituting these coordinates into the normal equation, we obtain:

$$(0 - y_1) = \frac{y_1}{2x_1}\left(-\frac{1}{3\sqrt{2}} - x_1ight)$$

Since $P$ is in the first quadrant, $y_1 
eq 0$. We can divide both sides by $y_1$ to simplify the equation:

$$-1 = \frac{1}{2x_1}\left(-\frac{1}{3\sqrt{2}} - x_1ight)$$

Solving for $x_1$, we find:

$$-2x_1 = -\frac{1}{3\sqrt{2}} - x_1 \Rightarrow x_1 = \frac{1}{3\sqrt{2}}$$

### The Final Calculation

With $x_1$ determined, we find $y_1$ by substituting $x_1$ back into the original ellipse equation:

$$2\left(\frac{1}{3\sqrt{2}}ight)^2 + y_1^2 = 1 \Rightarrow 2\left(\frac{1}{18}ight) + y_1^2 = 1$$

This simplifies to $\frac{1}{9} + y_1^2 = 1$, which gives $y_1^2 = \frac{8}{9}$. Given that $P$ is in the first quadrant:

$$y_1 = \frac{2\sqrt{2}}{3}$$

To find the $y$-intercept $\beta$, we set $x=0$ and $y=\beta$ in the normal equation:

$$(\beta - y_1) = \frac{y_1}{2x_1}(0 - x_1)$$

The $x_1$ terms cancel out, resulting in $\beta - y_1 = -\frac{y_1}{2}$, which simplifies to $\beta = \frac{y_1}{2}$. Substituting the value of $y_1$:

$$\beta = \frac{1}{2} 	imes \frac{2\sqrt{2}}{3} = \frac{\sqrt{2}}{3}$$

The final $y$-intercept is **$\frac{\sqrt{2}}{3}$**.

Animated Solution for Mathematics - Conic Sections: Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ intersect a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(- \frac{1}{3 2}, 0)$ and $(0, β)$ , then $β$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Setup

The Normal Line and Intercepts

Slope of the Tangent

Slope of the Normal

Equation of the Normal

Using the $x$ -intercept

Solving for $x_{1}$

Calculating $x_{1}$

Finding $y_{1}$ using the Ellipse Equation

Calculating $y_{1}$

Using the $y$ -intercept

Simplifying for $β$

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Anatomy of the Normal

Building the Bridge

The Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Setup

The Normal Line and Intercepts

Slope of the Tangent

Slope of the Normal

Equation of the Normal

Using the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-intercept

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y1​ using the Ellipse Equation

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y1​

Using the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-intercept

Simplifying for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β