The normal at a point P on the ellipse x² + 4y² = 16 meets the x-axis at Q. If M is the mid point of

Q: Why do we use parametric coordinates? Can't we just use $y = mx + c$?

You could, but it would lead to a nightmare of algebraic manipulation. Parametric coordinates $P(a\cos\theta, b\sin\theta)$ simplify the normal equation significantly, turning a complex geometry problem into a simple trigonometric identity problem.

Q: Why is the normal equation $ax\sec\theta - by\csc\theta = a^2 - b^2$?

This is a standard result derived from the slope of the tangent. The tangent at $\theta$ has slope $-\frac{b}{a}\cot\theta$. The normal, being perpendicular, has slope $\frac{a}{b}\tan\theta$. Using the point-slope form gives you this elegant equation.

Q: How do I know the locus is an ellipse?

Look at the final equation $\frac{4h^2}{49} + k^2 = 1$. It matches the form $\frac{X^2}{A^2} + \frac{Y^2}{B^2} = 1$. Since the coefficients of $h^2$ and $k^2$ are positive and unequal, it is indeed an ellipse.

Question

The normal at a point $P$ on the ellipse $x^2 + 4y^2 = 16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $PQ$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points

Options:

(A) $(\pm \frac{3\sqrt{5}}{2}, \pm \frac{2}{7})$

(B) $(\pm \frac{3\sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4})$

(C) $(\pm 2\sqrt{3}, \pm \frac{1}{7})$

(D) $(\pm 2\sqrt{3}, \pm \frac{4\sqrt{3}}{7})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given ellipse is defined by the equation $x^2 + 4y^2 = 16$. To bring this into the standard form, we divide the entire equation by $16$:

$$\frac{x^2}{16} + \frac{y^2}{4} = 1$$

Here, the semi-major axis is $a = 4$ and the semi-minor axis is $b = 2$. This serves as the foundation for our geometric analysis.

### The Normal's Path

Consider a point $P$ on the ellipse. We utilize the parametric form $P(4\cos	heta, 2\sin	heta)$ to simplify our calculations. The standard equation for a normal to an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a\cos	heta, b\sin	heta)$ is given by:

$$ax\sec	heta - by\csc	heta = a^2 - b^2$$

Substituting our specific values $a=4$ and $b=2$, the equation of the normal becomes:

$$4x\sec	heta - 2y\csc	heta = 12$$

### The Intersection and the Midpoint

The normal meets the $x$-axis at point $Q$. Setting $y=0$ in the normal equation, we find $4x\sec	heta = 12$, which simplifies to $x = 3\cos	heta$. Thus, the coordinates of $Q$ are $(3\cos	heta, 0)$.

We now seek the midpoint $M(h, k)$ of the segment $PQ$. Using the midpoint formula, we calculate the coordinates as follows:

$$h = \frac{4\cos	heta + 3\cos	heta}{2} = \frac{7}{2}\cos	heta$$

$$k = \frac{2\sin	heta + 0}{2} = \sin	heta$$

### The Locus Revealed

To find the locus of $M$, we eliminate the parameter $	heta$. From our previous expressions, we have $\cos	heta = \frac{2h}{7}$ and $\sin	heta = k$. Applying the fundamental trigonometric identity $\cos^2	heta + \sin^2	heta = 1$, we arrive at the equation of the locus:

$$\frac{4h^2}{49} + k^2 = 1$$

### Final Calculation

We now determine the intersection of this locus with the latus rectums of the original ellipse. The eccentricity $e$ is given by $\sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}} = \frac{\sqrt{3}}{2}$. The latus rectum is located at $x = \pm ae = \pm 4 \cdot \frac{\sqrt{3}}{2} = \pm 2\sqrt{3}$.

Substituting $x = \pm 2\sqrt{3}$ into our locus equation:

$$\frac{4(12)}{49} + y^2 = 1 \Rightarrow y^2 = 1 - \frac{48}{49} = \frac{1}{49}$$

Solving for $y$, we find $y = \pm \frac{1}{7}$. The intersection points are **$(\pm 2\sqrt{3}, \pm \frac{1}{7})$**.

Animated Solution for Mathematics - Conic Sections: The normal at a point $P$ on the ellipse $x^{2} + 4 y^{2} = 16$ meets the $x$ -axis at $Q$ . If $M$ is the mid point of the line segment $P Q$ , then the locus of $M$ intersects the latus rectums of the given ellipse at the points

Select Answer:

Visualized Solution

Standard Form of Ellipse

Eccentricity & Latus Rectum

Point $P$ and Normal Equation

Normal Equation Substitution

Finding Point $Q$

Midpoint $M (h, k)$

Locus of $M$

Intersection with Latus Rectum

Solving for $y$

Final Coordinates

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Normal's Path

The Intersection and the Midpoint

The Locus Revealed

Final Calculation

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Select Answer:

Visualized Solution

Standard Form of Ellipse

Eccentricity & Latus Rectum

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

Normal Equation Substitution

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

Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M(h,k)

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