Let P be a parabola with vertex (2, 3) and directrix 2x + y = 6. Let an ellipse E : x²/a² + y²/b² =

Q: Wait, why is the axis of the parabola perpendicular to the directrix?

That is a great question! The axis of symmetry is defined as the line passing through the focus and perpendicular to the directrix. By the very definition of a parabola, every point on the curve is equidistant from the focus and the directrix, which forces the curve to be symmetric about this perpendicular line.

Q: Can we use the definition of a parabola (distance from focus = distance from directrix) instead of the midpoint property?

Yes, you absolutely could! You could set up the distance formula $PF = PM$, but the midpoint property is a much faster, elegant shortcut often used in JEE problems to save precious time. It is a powerful tool to keep in your arsenal.

Q: Why is $a^2 = 2b^2$ in this ellipse?

The eccentricity $e$ is given by the relation $e^2 = 1 - \frac{b^2}{a^2}$. Substituting the given $e = \frac{1}{\sqrt{2}}$, we get $e^2 = \frac{1}{2}$. So, $\frac{1}{2} = 1 - \frac{b^2}{a^2}$, which simplifies to $\frac{b^2}{a^2} = \frac{1}{2}$, or $a^2 = 2b^2$. It is just a direct consequence of the definition of eccentricity!

Question

Let $P$ be a parabola with vertex $(2, 3)$ and directrix $2x + y = 6$. Let an ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ of eccentricity $\frac{1}{\sqrt{2}}$ pass through the focus of the parabola $P$. Then the square of the length of the latus rectum of $E$, is

Options:

(A) $\frac{385}{8}$

(B) $\frac{347}{8}$

(C) $\frac{512}{25}$

(D) $\frac{656}{25}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of a parabola is defined by the relationship between its focus and its directrix. We are given the vertex $V(2, 3)$ and the directrix line $2x + y - 6 = 0$.

Our objective is to determine the focus $F$ of this parabola, which subsequently serves as a point on a specific ellipse.

### Unlocking the Parabola

The axis of symmetry is perpendicular to the directrix. Since the directrix has a slope of $-2$, the axis must have a slope of $\frac{1}{2}$.

Given that the axis passes through the vertex $V(2, 3)$, its equation is:
$$y - 3 = \frac{1}{2}(x - 2) \Rightarrow x - 2y + 4 = 0$$

Next, we find the intersection point $M$ of the axis and the directrix by solving the system:
$$2x + y = 6$$
$$x - 2y = -4$$

Solving these simultaneously yields the intersection point $M = (1.6, 2.8)$.

### The Midpoint Magic

The vertex $V$ is the midpoint of the segment connecting the focus $F$ and the intersection point $M$. Using the midpoint formula $V = \frac{F+M}{2}$, we isolate the focus:
$$F = 2V - M$$

Substituting the coordinates $V(2, 3)$ and $M(1.6, 2.8)$:
$$x_f = 2(2) - 1.6 = 2.4$$
$$y_f = 2(3) - 2.8 = 3.2$$

Thus, the focus is $F(2.4, 3.2)$.

### The Ellipse Encounter

We consider the ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity $e = \frac{1}{\sqrt{2}}$. Using the relation $e^2 = 1 - \frac{b^2}{a^2}$, we find:
$$\frac{1}{2} = 1 - \frac{b^2}{a^2} \Rightarrow a^2 = 2b^2$$

Since the ellipse passes through $F(2.4, 3.2)$, we substitute these coordinates into the ellipse equation:
$$\frac{(2.4)^2}{2b^2} + \frac{(3.2)^2}{b^2} = 1$$

Simplifying the expression:
$$\frac{5.76}{2b^2} + \frac{10.24}{b^2} = 1 \Rightarrow \frac{2.88 + 10.24}{b^2} = 1$$
$$\frac{13.12}{b^2} = 1 \Rightarrow b^2 = 13.12 = \frac{328}{25}$$

### Final Calculation

The length of the latus rectum $L$ is given by $L = \frac{2b^2}{a}$. We are asked to find the square of this length, $L^2 = \frac{4b^4}{a^2}$.

Substituting $a^2 = 2b^2$ into the equation:
$$L^2 = \frac{4b^4}{2b^2} = 2b^2$$

Finally, substituting the value of $b^2$:
$$L^2 = 2 	imes \frac{328}{25} = \frac{656}{25}$$

The final result is **$\frac{656}{25}$** (or **$26.24$**).

Select Answer:

Visualized Solution

Identify Parabola Parameters

Equation of the Axis

Locate Point $M$ on Directrix

Calculate Coordinates of $M$

Apply Midpoint Theorem for Focus

Calculate Focus $F$

Ellipse Eccentricity Relation

Substitute Focus into Ellipse

Solve for $b^{2}$

Length of Latus Rectum

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Unlocking the Parabola

The Midpoint Magic

The Ellipse Encounter

Final Calculation

Similar Questions

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If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, - 3)$ and eccentricity is $\frac{4}{5}$ , is

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Select Answer:

Visualized Solution

Identify Parabola Parameters

Equation of the Axis

Locate Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M on Directrix

Calculate Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Apply Midpoint Theorem for Focus

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

Ellipse Eccentricity Relation

Substitute Focus into Ellipse

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Length of Latus Rectum

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Unlocking the Parabola

The Midpoint Magic

The Ellipse Encounter

Final Calculation

Similar Questions

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