A normal with slope 1/√6 is drawn from the point (0, -a) to the parabola x² = -4ay, where a 0. Let

Q: Wait, why do we use the parametric form $P(2at, -at^2)$?

Great question! Using the parametric form is a powerful technique in coordinate geometry. For the parabola $x^2 = -4ay$, any point on it can be represented as $(2at, -at^2)$. This reduces the number of variables we have to deal with from two ($x$ and $y$) to just one ($t$), making the differentiation and slope calculations much cleaner.

Q: How do we know the directrix is $y=a$?

For a standard parabola $x^2 = 4ay$, the directrix is $y = -a$. Here, our equation is $x^2 = -4ay$. By comparing this to the standard form $x^2 = 4Ay$, we see that $A = -a$. Thus, the directrix is $y = -A$, which becomes $y = -(-a) = a$. It is a simple sign check that prevents a major error!

Q: Why is the length of the latus rectum $4a$?

The length of the latus rectum for any parabola $x^2 = 4Ay$ or $x^2 = -4Ay$ is defined as the absolute value of the coefficient of the linear term, which is $|4A|$. In our case, the equation is $x^2 = -4ay$, so the length is $|-4a| = 4a$. It is a fundamental property of parabolas that you should always keep in your toolkit.

Question

A normal with slope $\frac{1}{\sqrt{6}}$ is drawn from the point $(0, -a)$ to the parabola $x^2 = -4ay$, where $a > 0$. Let $L$ be the line passing through $(0, -a)$ and parallel to the directrix of the parabola. Suppose that $L$ intersects the parabola at two points $A$ and $B$. Let $r$ denote the length of the latus rectum and $s$ denote the square of the length of the line segment $AB$. If $r : s = 1 : 16$, then the value of $24a$ is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The parabola is defined by the equation $x^2 = -4ay$. This represents a downward-opening parabola with its vertex at the origin.

To navigate this curve, we utilize the parametric form. A point $P$ on the parabola is represented as $P(2at, -at^2)$, where $t$ is the parameter.

### The Parametric Dance

To find the slope of the tangent, we differentiate $x^2 = -4ay$ with respect to $x$:

$$2x = -4a \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = -\frac{x}{2a}$$

At the point $P(2at, -at^2)$, the slope of the tangent $m_T$ is:

$$m_T = -\frac{2at}{2a} = -t$$

Since the normal is perpendicular to the tangent, its slope $m_N$ is the negative reciprocal:

$$m_N = \frac{1}{t}$$

### The Normal's Path

Given that the slope of the normal is $\frac{1}{\sqrt{6}}$, we equate this to our derived slope:

$$\frac{1}{t} = \frac{1}{\sqrt{6}} \Rightarrow t = \sqrt{6}$$

Substituting $t$ back into the coordinates of $P$, we find $P(2\sqrt{6}a, -6a)$. The equation of the normal line in point-slope form is:

$$y - (-6a) = \frac{1}{\sqrt{6}}(x - 2\sqrt{6}a)$$

Simplifying this expression yields:

$$y + 6a = \frac{x}{\sqrt{6}} - 2a \Rightarrow y = \frac{x}{\sqrt{6}} - 8a$$

This line intersects the $y$-axis at the point $(0, -8a)$.

### The Line $L$ and the Intersection

The directrix of the parabola $x^2 = -4ay$ is the horizontal line $y = a$. Since line $L$ is parallel to the directrix and passes through $(0, -8a)$, its equation is $y = -8a$.

To find the intersection points of $L$ and the parabola, we substitute $y = -8a$ into $x^2 = -4ay$:

$$x^2 = -4a(-8a) = 32a^2$$

This results in $x = \pm 4\sqrt{2}a$. Thus, the intersection points are $A(4\sqrt{2}a, -8a)$ and $B(-4\sqrt{2}a, -8a)$.

The length of the segment $AB$ is the difference in the $x$-coordinates:

$$|AB| = 4\sqrt{2}a - (-4\sqrt{2}a) = 8\sqrt{2}a$$

The square of this length, $s$, is:

$$s = (8\sqrt{2}a)^2 = 128a^2$$

### Final Calculation

The length of the latus rectum $r$ for the parabola $x^2 = -4ay$ is $4a$. We are given the ratio $r : s = 1 : 16$:

$$\frac{4a}{128a^2} = \frac{1}{16}$$

Simplifying the left side:

$$\frac{1}{32a} = \frac{1}{16} \Rightarrow 32a = 16 \Rightarrow a = \frac{1}{2}$$

The final value requested is $24a$:

$$24a = 24 	imes \frac{1}{2} = \mathbf{12}$$

Enter Numerical Value:

Visualized Solution

Parabola Equation

Slopes of Tangent and Normal

Finding Parameter $t$

Equation of Normal

$y$ -intercept of Normal

Identifying Line $L$

Equation of Line $L$

Intersection Setup

Coordinates of $A$ and $B$

Length of Segment $A B$

Latus Rectum Length

Using the Given Ratio

Solving for $a$

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Dance

The Normal's Path

The Line $L$ and the Intersection

Final Calculation

Similar Questions

Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

$A$ is a point on the parabola $y^{2} = 4 a x$ . The normal at $A$ cuts the parabola again at point $B$ . If $A B$ subtends a right angle at the vertex of the parabola, find the slope of $A B$ .

Comprehension Passage

The value of $r$ is

If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ , then the value of $r^{2}$ is

A tangent and a normal are drawn at the point $P (2, - 4)$ on the parabola $y^{2} = 8 x$ , which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q (a, b)$ is a point such that $A QB P$ is a square, then $2 a + b$ is equal to :

If the three normals drawn to the parabola, $y^{2} = 2 x$ pass through the point $(a, 0) a \neq = 0$ , then 'a' must be greater than :

Consider the parabola with vertex $(\frac{1}{2}, \frac{3}{4})$ and the directrix $y = \frac{1}{2}$ . Let $P$ be the point where the parabola meets the line $x = - \frac{1}{2}$ . If the normal to the parabola at $P$ intersects the parabola again at the point $Q$ , then $(P Q)^{2}$ is equal to :

The tangent $P T$ and the normal $P N$ to the parabola $y^{2} = 4 a x$ at a point $P$ on it meet its axis at points $T$ and $N$ , respectively. The locus of the centroid of the triangle $P T N$ is a parabola whose

Enter Numerical Value:

Visualized Solution

Parabola Equation

Slopes of Tangent and Normal

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

Equation of Normal

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

Identifying Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

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

Intersection Setup

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

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

Latus Rectum Length

Using the Given Ratio

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

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Dance

The Normal's Path

The Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L and the Intersection

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Normals are drawn from the point P with slopes m1​,m2​,m3​ to the parabola y2=4x. If locus of P with m1​m2​=α is a part of the parabola itself then find α.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A is a point on the parabola y2=4ax. The normal at A cuts the parabola again at point B. If AB subtends a right angle at the vertex of the parabola, find the slope of AB.

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of r is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If st=1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normals of the parabola y2=4x drawn at the end points of its latus rectum are tangents to the circle (x−3)2+(y+2)2=r2, then the value of r2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the three normals drawn to the parabola, y2=2x pass through the point (a,0)a=0, then 'a' must be greater than :

Finding Parameter $t$

$y$ -intercept of Normal

Identifying Line $L$

Equation of Line $L$

Coordinates of $A$ and $B$

Length of Segment $A B$

Solving for $a$

The Line $L$ and the Intersection

Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

$A$ is a point on the parabola $y^{2} = 4 a x$ . The normal at $A$ cuts the parabola again at point $B$ . If $A B$ subtends a right angle at the vertex of the parabola, find the slope of $A B$ .

The value of $r$ is

If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ , then the value of $r^{2}$ is

If the three normals drawn to the parabola, $y^{2} = 2 x$ pass through the point $(a, 0) a \neq = 0$ , then 'a' must be greater than :