JEE Advanced 2004

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Tangent is drawn to parabola $y^{2} - 2 y - 4 x + 5 = 0$ at a point $P$ which cuts the directrix at the point $Q$ . A point $R$ is such that it divides $QP$ externally in the ratio $1/2 : 1$ . Find the locus of point $R$ .

Q: I found the coordinates of $R$ in terms of $t$, but how do I know which variable to eliminate first?

Look at your expressions for $h$ and $k$. Usually, one coordinate will be linear in $t$ (or easily invertible), and the other will be quadratic. Always isolate $t$ in the simpler, linear-like equation first. Once you have $t$ in terms of $k$, substitute that into the quadratic equation for $h$. It is a systematic process of 'parameter hunting'—isolate, substitute, and simplify.

Progressive Hints

Hint 1

Don't let the shifted equation intimidate you. The first step to mastering any conic section is to find its true center. Complete the square to bring the parabola into its standard form

(y - 1)^{2} = 4 (x - 1)

. Visualizing the vertex at

(1, 1)

makes the entire problem collapse into a much simpler, symmetric world.

Hint 2

Now that you have the standard form

Y^{2} = 4 a X

, where

a = 1

, you need the parametric toolkit. Remember that any point

P

on this parabola can be elegantly represented as

(t^{2} + 1, 2 t + 1)

. The tangent at this point is not just a line; it is a geometric constraint defined by

t Y = X + a t^{2}

. Use this to find the intersection with the directrix.

Hint 3

The final hurdle is the locus of

R

. You are given that

R

divides

QP

externally in the ratio

1/2 : 1

. This is equivalent to a ratio of

1 : 2

. Recall the external section formula:

R = \frac{m P - n Q}{m - n}

. With

m = 1

and

n = 2

, this simplifies beautifully to

R = 2 Q - P

. Now, express the coordinates of

R

in terms of

t

and eliminate the parameter.

Visualized Solution

Standard Form of the Parabola

Given equation: $y^{2} - 2 y - 4 x + 5 = 0$
Completing the square for $y$ : $(y^{2} - 2 y + 1) - 1 - 4 x + 5 = 0$
$(y - 1)^{2} = 4 x - 4$
Standard Form: $(y - 1)^{2} = 4 (x - 1)$

Parametric Coordinates of Point $P$

Let $X = x - 1$ and $Y = y - 1$ . The parabola is $Y^{2} = 4 X$ .
Parametric form: $X = t^{2}, Y = 2 t$ (since $a = 1$ )
Substituting back: $x - 1 = t^{2} ⟹ x = t^{2} + 1$
$y - 1 = 2 t ⟹ y = 2 t + 1$
Point $P = (t^{2} + 1, 2 t + 1)$

Equation of the Tangent at $P$

Tangent at $(X_{1}, Y_{1})$ is $Y Y_{1} = 2 a (X + X_{1})$ or $t Y = X + a t^{2}$
Substituting $X = x - 1$ and $Y = y - 1$ : $t (y - 1) = (x - 1) + t^{2}$
$t y - t = x - 1 + t^{2}$
Equation of tangent: $x - t y + t^{2} + t - 1 = 0$

Finding Point $Q$ on the Directrix

Directrix of $(y - 1)^{2} = 4 (x - 1)$ is $x - 1 = - 1 ⟹ x = 0$
Intersection of tangent and $x = 0$ :
$0 - t y + t^{2} + t - 1 = 0 ⟹ t y = t^{2} + t - 1$
$y_{Q} = \frac{t ^{2} + t - 1}{t} = t + 1 - \frac{1}{t}$
Point $Q = (0, t + 1 - \frac{1}{t})$

External Division for Point $R$

Ratio $m : n = \frac{1}{2} : 1 = 1 : 2$ externally.
Section formula: $R = \frac{m P - n Q}{m - n} = \frac{1 P - 2 Q}{1 - 2} = 2 Q - P$
$h = 2 (0) - (t^{2} + 1) = - t^{2} - 1$
$k = 2 (t + 1 - \frac{1}{t}) - (2 t + 1) = 1 - \frac{2}{t}$

Eliminating the Parameter $t$

From $k = 1 - \frac{2}{t} ⟹ k - 1 = - \frac{2}{t} ⟹ t = \frac{2}{1 - k}$
From $h = - t^{2} - 1 ⟹ h + 1 = - t^{2}$
Substitute $t$ : $h + 1 = - (\frac{2}{1 - k})^{2}$

Final Locus of Point $R$

$h + 1 = - \frac{4}{( k - 1 ) ^{2}}$
$(h + 1) (k - 1)^{2} = - 4$
$(h + 1) (k - 1)^{2} + 4 = 0$
Replacing $(h, k)$ with $(x, y)$ :
Final Locus: $(x + 1) (y - 1)^{2} + 4 = 0$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We begin with the equation

y^{2} - 2 y - 4 x + 5 = 0

. By completing the square for the

y

terms, we transform this into:

(y - 1)^{2} = 4 (x - 1)

This represents a standard parabola

Y^{2} = 4 X

shifted to a vertex at

(1, 1)

. Recognizing this form is the first step in simplifying the geometry of the problem.

The Parametric Dance

For the parabola

Y^{2} = 4 X

(where

a = 1

), any point

P

can be defined using the parameter

t

as:

P = (t^{2} + 1, 2 t + 1)

The equation of a tangent to a parabola

Y^{2} = 4 a X

at point

t

is given by

t Y = X + a t^{2}

. Substituting our shifted variables back in, we derive the equation of the tangent:

x - t y + t^{2} + t - 1 = 0

The Intersection at the Directrix

The directrix of our parabola is

x - 1 = - a

, which simplifies to

x - 1 = - 1

, or

x = 0

. This is the

y

-axis.

By setting

x = 0

in our tangent equation, we find the intersection point

Q

0 - t y + t^{2} + t - 1 = 0 \Rightarrow y_{Q} = t + 1 - \frac{1}{t}

Thus, the coordinates of point

Q

are

(0, t + 1 - \frac{1}{t})

. We have now defined both

P

and

Q

in terms of the parameter

t

The Locus of R

The Final Act

Point

R

divides

QP

externally in the ratio

1 : 2

. Using the external section formula,

R = \frac{1 \cdot P - 2 \cdot Q}{1 - 2} = 2 Q - P

Calculating the coordinates

(h, k)

R

h = 2 (0) - (t^{2} + 1) = - t^{2} - 1

k = 2 (t + 1 - \frac{1}{t}) - (2 t + 1) = 1 - \frac{2}{t}

To find the locus, we eliminate

t

. From

k = 1 - \frac{2}{t}

, we isolate

t

t = \frac{2}{1 - k}

Substituting this into the expression for

h

h + 1 = - (\frac{2}{1 - k})^{2}

Rearranging the terms, we arrive at the final locus:

(x + 1) (y - 1)^{2} + 4 = 0

This equation represents the geometric signature of the point

R

. The final locus is $(x + 1) (y - 1)^{2} + 4 = 0$ .

Common Traps & FAQs

Why do we bother shifting the origin? Can't we just use the original equation

y^{2} - 2 y - 4 x + 5 = 0

You absolutely could, but you would be fighting a war on two fronts: the geometry and the messy algebra. By shifting the origin to

(1, 1)

, we transform the parabola into the standard

Y^{2} = 4 X

. This allows us to use the standard parametric forms we have memorized, drastically reducing the probability of a sign error. In JEE Advanced, elegance is not just for show; it is a survival strategy.

The ratio is

1/2 : 1

. How do I handle this in the section formula without getting confused?

Great question! Never work with fractions in a ratio if you can avoid it. Multiply both sides by 2 to get the integer ratio

1 : 2

. When applying the external division formula

\frac{m P - n Q}{m - n}

, remember that the negative sign is already accounted for in the formula. If you use

m = 1

and

n = 2

, the denominator becomes

1 - 2 = - 1

, which effectively flips the signs of your numerator. Always double-check your signs!

I found the coordinates of

R

in terms of

t

, but how do I know which variable to eliminate first?

Look at your expressions for

h

and

k

. Usually, one coordinate will be linear in

t

(or easily invertible), and the other will be quadratic. Always isolate

t

in the simpler, linear-like equation first. Once you have

t

in terms of

k

, substitute that into the quadratic equation for

h

. It is a systematic process of 'parameter hunting'—isolate, substitute, and simplify.

Animated Solution for Mathematics - Conic Sections: Tangent is drawn to parabola $y^{2} - 2 y - 4 x + 5 = 0$ at a point $P$ which cuts the directrix at the point $Q$ . A point $R$ is such that it divides $QP$ externally in the ratio $1/2 : 1$ . Find the locus of point $R$ .

Visualized Solution

Standard Form of the Parabola

Parametric Coordinates of Point $P$

Equation of the Tangent at $P$

Finding Point $Q$ on the Directrix

External Division for Point $R$

Eliminating the Parameter $t$

Final Locus of Point $R$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Dance

The Intersection at the Directrix

The Locus of R

Similar Questions

If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

Let $C$ be the locus of the mirror image of a point on the parabola $y^{2} = 4 x$ with respect to the line $y = x$ . Then the equation of tangent to $C$ at $P (2, 1)$ 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

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 $R (α, β)$ is the point of intersection of the tangents to the parabola at $P$ and $Q$ , then $\frac{β ^{2}}{α}$ is equal to

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

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 $α$ .

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 :

Visualized Solution

Standard Form of the Parabola

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

Equation of the Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

External Division for Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

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

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

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Dance

The Intersection at the Directrix

The Locus of R

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=4x are at right angles, then the locus of P is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between a pair of tangents drawn from a point P to the parabola y2=4ax is 45∘. Show that the locus of the point P is a hyperbola.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let C be the locus of the mirror image of a point on the parabola y2=4x with respect to the line y=x. Then the equation of tangent to C at P(2,1) 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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

.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; } 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 :

Parametric Coordinates of Point $P$

Equation of the Tangent at $P$

Finding Point $Q$ on the Directrix

External Division for Point $R$

Eliminating the Parameter $t$

Final Locus of Point $R$

If two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

Let $C$ be the locus of the mirror image of a point on the parabola $y^{2} = 4 x$ with respect to the line $y = x$ . Then the equation of tangent to $C$ at $P (2, 1)$ is :

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 a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

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 $α$ .

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 :