If the normal to the ellipse 3x² + 4y² = 12 at a point P on it is parallel to the line, 2x + y = 4 a

Q: Wait, why do we use the slope of the normal instead of the tangent?

Great question! The problem explicitly states that the *normal* is parallel to the line $2x + y = 4$. If the normal is parallel to this line, it must have the same slope as the line. If we had used the tangent, we would have been looking for a line perpendicular to the given line, which would lead us to the wrong point $P$.

Q: How do I know which value of $\theta$ to choose if there are two possibilities?

Excellent observation. When you solve $\tan\theta = -\sqrt{3}$, you get two angles in the range $[0, 2\pi)$. However, the problem provides an additional constraint: the tangent at $P$ must pass through $Q(4, 4)$. You should test both potential points $P$ by finding the tangent equation at each and checking if $Q$ satisfies it. Only the correct point will satisfy this condition.

Q: Is there a faster way to find the distance $PQ$ without finding the coordinates of $P$?

While you could potentially use properties of the ellipse, finding the coordinates of $P$ is the most robust and reliable method for this type of problem. It ensures you satisfy all conditions (normal slope and tangent passing through $Q$) simultaneously. In JEE Advanced, accuracy is just as important as speed!

Question

If the normal to the ellipse $3x^2 + 4y^2 = 12$ at a point P on it is parallel to the line, $2x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

Options:

(A) $\sqrt{221}/2$

(B) $\sqrt{157}/2$

(C) $\sqrt{61}/2$

(D) $5\sqrt{5}/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the ellipse equation $3x^2 + 4y^2 = 12$. To analyze this geometry, we first convert it to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Dividing the entire equation by $12$, we obtain:

$$ \frac{x^2}{4} + \frac{y^2}{3} = 1 $$

From this, we identify the semi-major axis $a = 2$ and the semi-minor axis $b = \sqrt{3}$.

### The Parametric Elegance

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

$$ P = (2\cos	heta, \sqrt{3}\sin	heta) $$

This trigonometric substitution allows us to handle the slope conditions of the normal line with greater ease.

### The Normal's Path

The problem states that the normal at $P$ is parallel to the line $2x + y = 4$. Rewriting the line as $y = -2x + 4$, we identify its slope as $m = -2$.

The slope of the normal $m_n$ at a point $(x_1, y_1)$ on the ellipse is given by:

$$ m_n = \frac{a^2 y_1}{b^2 x_1} $$

Substituting our parametric coordinates $x_1 = 2\cos	heta$ and $y_1 = \sqrt{3}\sin	heta$, we get:

$$ m_n = \frac{4(\sqrt{3}\sin	heta)}{3(2\cos	heta)} = \frac{2}{\sqrt{3}}	an	heta $$

Equating this to the slope of the given line, we have $\frac{2}{\sqrt{3}}	an	heta = -2$, which simplifies to $	an	heta = -\sqrt{3}$.

### Finding $P$ and Verifying the Tangent

Given $	an	heta = -\sqrt{3}$, we consider $	heta = 120^\circ$. The coordinates of point $P$ are:

$$ P = (2\cos(120^\circ), \sqrt{3}\sin(120^\circ)) = \left(-1, \frac{3}{2}ight) $$

The equation of the tangent at $P(x_1, y_1)$ is $\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$. Substituting our values:

$$ \frac{x(-1)}{4} + \frac{y(3/2)}{3} = 1 \implies -\frac{x}{4} + \frac{y}{2} = 1 \implies -x + 2y = 4 $$

Testing the point $Q(4, 4)$ in this tangent equation: $-4 + 2(4) = 4$. Since the condition is satisfied, our point $P$ is correct.

### Final Calculation

We now calculate the distance $PQ$ between $P(-1, \frac{3}{2})$ and $Q(4, 4)$ using the distance formula:

$$ PQ = \sqrt{(4 - (-1))^2 + \left(4 - \frac{3}{2}ight)^2} $$

$$ PQ = \sqrt{5^2 + \left(\frac{5}{2}ight)^2} = \sqrt{25 + \frac{25}{4}} = \sqrt{\frac{125}{4}} $$

The final distance is:

$$ \mathbf{PQ = \frac{5\sqrt{5}}{2}} $$

Animated Solution for Mathematics - Conic Sections: If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point P on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

Select Answer:

Visualized Solution

Standardizing the Ellipse Equation

Defining Point $P$ Parametrically

Finding the Slope of the Normal

Analyzing the Given Line

Solving for the Parameter $θ$

Determining Coordinates of Point $P$

Writing the Tangent Equation

Verifying the Tangent Condition

Setting up the Distance Formula

Calculating the Final Distance $P Q$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Elegance

The Normal's Path

Finding $P$ and Verifying the Tangent

Final Calculation

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Select Answer:

Visualized Solution

Standardizing the Ellipse Equation

Defining Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P Parametrically

Finding the Slope of the Normal

Analyzing the Given Line

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

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