The ordinates of the points P and Q on the parabola with focus (3, 0) and directrix x = -3 are in th

Q: Wait, why is the vertex at the origin? How do I know that?

Great question! The vertex of a parabola is always the midpoint between the focus and the directrix. Since the focus is at $(3, 0)$ and the directrix is the line $x = -3$, the midpoint is $(\frac{3 + (-3)}{2}, 0)$, which is $(0, 0)$. That is our vertex!

Q: Why can't we just use the slope-intercept form of the tangent?

You absolutely can! However, using parametric coordinates $(at^2, 2at)$ is often much faster for JEE problems involving tangents on parabolas. It turns complex algebraic expressions into simple polynomial manipulations, saving you precious time during the exam.

Q: Why does the $t_2^2$ term cancel out in the final step?

It cancels because the ratio of the ordinates is a constant. When you express $\beta^2$ and $\alpha$ in terms of $t_2$, both end up being proportional to $t_2^2$. This is a beautiful property of the geometry of the parabola—the ratio of the squared ordinate to the abscissa of the intersection of tangents depends only on the ratio of the parameters, not on the specific location of the points themselves.

Question

The ordinates of the points $P$ and $Q$ on the parabola with focus $(3, 0)$ and directrix $x = -3$ are in the ratio $3 : 1$. If $R(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $P$ and $Q$, then $\frac{\beta^2}{\alpha}$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, empty plane. You have a fixed point, the focus at $(3, 0)$, and a rigid boundary, the directrix at $x = -3$.

A parabola is the set of all points equidistant from this focus and this directrix. Because the focus and directrix are perfectly symmetric about the $y$-axis, the vertex of our parabola must lie exactly in the middle, at the origin $(0, 0)$.

The distance from the vertex to the focus is our parameter $a$, which is $3$. Thus, our parabola takes the standard form $y^2 = 4ax$, or more specifically:

$$y^2 = 12x$$

This is the stage upon which our geometric drama unfolds.

### The Parametric Dance

To analyze points on this curve, we use the power of parametric coordinates. Any point on the parabola $y^2 = 4ax$ can be elegantly described as $(at^2, 2at)$.

Let us place two points, $P$ and $Q$, on this curve. We assign $P$ the parameter $t_1$ and $Q$ the parameter $t_2$. Their coordinates are $P(3t_1^2, 6t_1)$ and $Q(3t_2^2, 6t_2)$.

The problem provides us with a crucial piece of information: the ratio of their ordinates (the $y$-coordinates) is $3:1$. Mathematically, this is:

$$\frac{6t_1}{6t_2} = 3$$

The $6$ cancels out, leaving us with the golden key: $t_1 = 3t_2$. This relationship will simplify our entire algebraic journey.

### The Intersection of Tangents

Now, imagine drawing the tangent lines at $P$ and $Q$. These lines are not parallel; they will eventually meet at a point $R(\alpha, \beta)$.

In the study of conics, the intersection of tangents at parameters $t_1$ and $t_2$ is a well-known result: the $x$-coordinate is $at_1t_2$ and the $y$-coordinate is $a(t_1 + t_2)$. Substituting our value of $a=3$, we find the coordinates of $R$ to be:

$$\alpha = 3t_1t_2$$
$$\beta = 3(t_1 + t_2)$$

### The Algebraic Simplification

We are now ready to bring our golden key, $t_1 = 3t_2$, into the mix. Let us substitute this into our expressions for $\alpha$ and $\beta$.

For $\alpha$, we have:

$$\alpha = 3(3t_2)(t_2) = 9t_2^2$$

For $\beta$, we have:

$$\beta = 3(3t_2 + t_2) = 3(4t_2) = 12t_2$$

We have successfully reduced the coordinates of the intersection point $R$ to depend solely on the parameter $t_2$.

### The Final Revelation

The problem asks us to evaluate the expression $\frac{\beta^2}{\alpha}$. With our expressions in hand, this becomes a simple calculation.

We square $\beta$ to get $\beta^2 = (12t_2)^2 = 144t_2^2$. Now, we divide this by $\alpha$:

$$\frac{\beta^2}{\alpha} = \frac{144t_2^2}{9t_2^2}$$

Notice the elegance of the result? The $t_2^2$ terms cancel out entirely, leaving us with $\frac{144}{9}$, which is exactly **16**.

This result is independent of the specific parameters $t_1$ and $t_2$, revealing a deep, underlying geometric truth about the tangents of this parabola. You have navigated the geometry, mastered the algebra, and arrived at the solution with precision.

Enter Numerical Value:

Visualized Solution

Identifying the Parabola $y^{2} = 12 x$

Defining Points $P$ and $Q$

Applying the Ratio $3 : 1$

Intersection of Tangents $R (α, β)$

Substituting $a = 3$ into $R$

Expressing $α$ in terms of $t_{2}$

Expressing $β$ in terms of $t_{2}$

Calculating $β^{2}$

Evaluating $\frac{β ^{2}}{α}$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Dance

The Intersection of Tangents

The Algebraic Simplification

The Final Revelation

Similar Questions

If two tangents drawn from a point $P$ to the parabola $y^{2} = 16 (x - 3)$ are at right angles, then the locus of point $P$ is :

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point:

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is

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

Comprehension Passage

Length of chord $P Q$ is

If chord $P Q$ subtends an angle $θ$ at the vertex of $y^{2} = 4 a x$ , then $tan θ =$

If the tangents drawn at the points $P$ and $Q$ on the parabola $y^{2} = 2 x - 3$ intersect at the point $R (0, 1)$ , then the orthocentre of the triangle $P QR$ is :

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

Let $P : y^{2} = 4 a x, a > 0$ be a parabola with focus $S$ . Let the tangents to the parabola $P$ make an angle of $\frac{π}{4}$ with the line $y = 3 x + 5$ touch the parabola $P$ at $A$ and $B$ . Then the value of $a$ for which $A, B$ and $S$ are collinear is:

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

Identifying the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=12x

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

Applying the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3:1

Intersection of Tangents .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R(α,β)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=3 into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2​

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β2

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } αβ2​

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Dance

The Intersection of Tangents

The Algebraic Simplification

The Final Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two tangents drawn from a point P to the parabola y2=16(x−3) are at right angles, then the locus of point P is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A hyperbola passes through the point P(2​,3​) and has foci at (±2,0). Then the tangent to this hyperbola at P also passes through the point:

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is

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

Comprehension Passage

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If chord PQ subtends an angle θ at the vertex of y2=4ax, then tanθ=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn at the points P and Q on the parabola y2=2x−3 intersect at the point R(0,1), then the orthocentre of the triangle PQR is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the parabola y2=x at a point (α,β),(β>0) is also a tangent to the ellipse, x2+2y2=1, then α is equal to :

Identifying the Parabola $y^{2} = 12 x$

Defining Points $P$ and $Q$

Applying the Ratio $3 : 1$

Intersection of Tangents $R (α, β)$

Substituting $a = 3$ into $R$

Expressing $α$ in terms of $t_{2}$

Expressing $β$ in terms of $t_{2}$

Calculating $β^{2}$

Evaluating $\frac{β ^{2}}{α}$

If two tangents drawn from a point $P$ to the parabola $y^{2} = 16 (x - 3)$ are at right angles, then the locus of point $P$ is :

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point:

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

Length of chord $P Q$ is

If chord $P Q$ subtends an angle $θ$ at the vertex of $y^{2} = 4 a x$ , then $tan θ =$

If the tangents drawn at the points $P$ and $Q$ on the parabola $y^{2} = 2 x - 3$ intersect at the point $R (0, 1)$ , then the orthocentre of the triangle $P QR$ is :

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :