Let the tangents at the points P and Q on the ellipse x²/2 + y²/4 = 1 meet at the point R(√2, 2√2 -

Q: Wait, why is the major axis vertical? I thought the major axis was always along the x-axis.

That is a common misconception! In the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the major axis is along the x-axis if $a^2 > b^2$. However, in our equation, the denominator under $y^2$ is $4$, which is greater than the denominator under $x^2$, which is $2$. This means the ellipse is stretched along the y-axis, making it a vertical ellipse.

Q: Is there a faster way to find $SP^2 + SQ^2$ without finding the coordinates of $P$ and $Q$?

While there are advanced properties involving focal distances, finding the coordinates of $P$ and $Q$ is the most robust and reliable method for this specific problem. The algebra is designed to simplify beautifully, so don't be intimidated by the substitution step!

Q: Why did we choose the negative focus $S(0, -\sqrt{2})$?

The problem explicitly specifies 'the focus of the ellipse on its negative major axis'. Since the major axis is the y-axis, the negative major axis corresponds to the negative y-axis. Therefore, we must select the focus with the negative y-coordinate.

Question

Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^2}{2} + \frac{y^2}{4} = 1$ meet at the point $R(\sqrt{2}, 2\sqrt{2} - 2)$. If $S$ is the focus of the ellipse on its negative major axis, then $SP^2 + SQ^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given equation of the ellipse is:

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

We observe that the denominator under $y^2$ is $4$, which is greater than the denominator under $x^2$ ($2$). This confirms that the ellipse is a **vertical ellipse** with its major axis along the $y$-axis.

### Finding the Heart of the Ellipse: The Focus

Every ellipse is defined by its foci. To find them, we first calculate the eccentricity $e$ using the formula for a vertical ellipse:

$$
e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}
$$

The foci are located at $(0, \pm be)$. Given $b = \sqrt{4} = 2$, the foci are at $(0, \pm 2 	imes \frac{1}{\sqrt{2}})$, which simplifies to $(0, \pm \sqrt{2})$.

The problem specifies the focus on the negative major axis. Therefore, we identify the point $S$ as:

$$
S = (0, -\sqrt{2})
$$

### The Bridge: The Chord of Contact

We are given an external point $R(\sqrt{2}, 2\sqrt{2} - 2)$. From this point, two tangents are drawn to the ellipse, touching at points $P$ and $Q$.

The chord of contact $PQ$ is found using the $T=0$ formula by substituting the coordinates of $R$ into the ellipse equation:

$$
\frac{x(\sqrt{2})}{2} + \frac{y(2\sqrt{2} - 2)}{4} = 1
$$

Simplifying this expression, we obtain the equation of the chord:

$$
\frac{x}{\sqrt{2}} + \frac{y(\sqrt{2} - 1)}{2} = 1
$$

### The Intersection: Solving for P and Q

To find the coordinates of $P$ and $Q$, we solve the chord equation simultaneously with the ellipse equation. Isolating $x$ from the chord equation gives:

$$
x = \sqrt{2} - \frac{y(\sqrt{2} - 1)}{\sqrt{2}}
$$

Substituting this into the ellipse equation $\frac{x^2}{2} + \frac{y^2}{4} = 1$ and simplifying the resulting quadratic in $y$, we find:

$$
y(\sqrt{2}-1) \left( \frac{y}{\sqrt{2}} - 1 ight) = 0
$$

This yields two solutions: $y=0$ and $y=\sqrt{2}$. Substituting these back into the linear equation, we determine the points:

$$
Q = (\sqrt{2}, 0), \quad P = (1, \sqrt{2})
$$

### The Final Calculation

We now calculate $SP^2 + SQ^2$ using the distance formula with $S(0, -\sqrt{2})$, $P(1, \sqrt{2})$, and $Q(\sqrt{2}, 0)$.

For $SP^2$:

$$
SP^2 = (1 - 0)^2 + (\sqrt{2} - (-\sqrt{2}))^2 = 1^2 + (2\sqrt{2})^2 = 1 + 8 = 9
$$

For $SQ^2$:

$$
SQ^2 = (\sqrt{2} - 0)^2 + (0 - (-\sqrt{2}))^2 = 2 + 2 = 4
$$

Summing these values, we reach the final result:

$$
SP^2 + SQ^2 = 9 + 4 = \mathbf{13}
$$

Animated Solution for Mathematics - Conic Sections: Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{4} = 1$ meet at the point $R (2, 22 - 2)$ . If $S$ is the focus of the ellipse on its negative major axis, then $S P^{2} + S Q^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Ellipse Equation

Finding Eccentricity $e$

Locating Focus $S$

Point $R$ and Chord of Contact

Simplifying the Chord Equation

Substitution into Ellipse

Expanding the Quadratic

Solving for $y$

Finding Point $Q$

Finding Point $P$

Calculating $S P^{2}$

Calculating $S Q^{2}$

Final Summation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Finding the Heart of the Ellipse

The Bridge

The Intersection

The Final Calculation

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Enter Numerical Value:

Visualized Solution

Analyzing the Ellipse Equation

Finding Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Locating Focus .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R and Chord of Contact