JEE Main 2021 (26 August Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: If a line along a chord of the circle $4 x^{2} + 4 y^{2} + 120 x + 675 = 0$ , passes through the point $(- 30, 0)$ and is tangent to the parabola $y^{2} = 30 x$ , then the length of this chord is :

Q: Wait, why do we get two slopes ($m = \pm 1/2$)? Does it matter which one I pick?

That is a fantastic observation! The point $P(-30, 0)$ lies outside the parabola $y^2 = 30x$. From any point outside a parabola, you can draw exactly two tangents. Both tangents will intersect the circle and form chords. However, due to the symmetry of the circle and the parabola about the x-axis, both chords will have the same length. So, picking either $m = 1/2$ or $m = -1/2$ will lead you to the same final result.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the geometry. You have a circle and a parabola. The line is a tangent to the parabola and a chord to the circle. The key to finding the chord length is to determine the perpendicular distance from the center of the circle to this line. Can you find the center and radius of the circle first?

Hint 2

Hint 2 (The Arsenal): Use the standard condition of tangency for a parabola

y^{2} = 4 a x

. If the line is

y = m x + c

, then

c = a / m

. Since the line passes through

P (- 30, 0)

, you can find the slope

m

. Once you have the equation of the line, use the perpendicular distance formula

p = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}

from the circle's center to the line.

Hint 3

Hint 3 (The Execution): The length of the chord is given by

2 R^{2} - p^{2}

. After finding the center

C (- 15, 0)

and radius

R = \frac{15}{2}

, and the line equation

x - 2 y + 30 = 0

, calculate

p

and substitute these values into the chord length formula.

Visualized Solution

Visualizing the Geometry

Given Circle: $4 x^{2} + 4 y^{2} + 120 x + 675 = 0$
Given Parabola: $y^{2} = 30 x$
Line passes through $P (- 30, 0)$ and is tangent to the parabola.
Goal: Find the length of the chord on the circle.

Analyzing the Circle

Divide circle equation by $4$ : $x^{2} + y^{2} + 30 x + \frac{675}{4} = 0$
Compare with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Center $C = (- g, - f) = (- 15, 0)$

Calculating the Radius

Radius $R = g^{2} + f^{2} - c$
$R = (- 15)^{2} + 0^{2} - \frac{675}{4}$
$R = 225 - \frac{675}{4} = \frac{900 - 675}{4}$
$R = \frac{225}{4} = \frac{15}{2}$

Tangent to the Parabola

Parabola: $y^{2} = 30 x \Rightarrow 4 a = 30 \Rightarrow a = \frac{15}{2}$
Equation of tangent with slope $m$ : $y = m x + \frac{a}{m}$
Substitute $a$ : $y = m x + \frac{15}{2 m}$

Finding the Slope $m$

Line passes through $P (- 30, 0)$ .
Substitute $x = - 30, y = 0$ : $0 = m (- 30) + \frac{15}{2 m}$
$30 m = \frac{15}{2 m} \Rightarrow 60 m^{2} = 15$
$m^{2} = \frac{15}{60} = \frac{1}{4} \Rightarrow m = \pm \frac{1}{2}$

Equation of the Line

Using $m = \frac{1}{2}$ : $y = \frac{1}{2} x + \frac{15}{2 ( 1/2 )}$
$y = \frac{1}{2} x + 15$
Multiply by $2$ : $2 y = x + 30$
General form: $x - 2 y + 30 = 0$

The Chord and Perpendicular Distance

We need the length of the chord formed by this line on the circle.
Formula: Length $= 2 R^{2} - p^{2}$
$p$ is the perpendicular distance from center $C (- 15, 0)$ to the line $x - 2 y + 30 = 0$ .

Calculating Perpendicular Distance $p$

$p = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$
$p = \frac{∣1 ( - 15 ) - 2 ( 0 ) + 30∣}{1 ^{2} + ( - 2 ) ^{2}}$
$p = \frac{∣ - 15 + 30∣}{1 + 4} = \frac{15}{5}$
$p = 35$

Calculating Chord Length

Substitute $R = \frac{15}{2}$ and $p = 35$ into the formula.
Length $= 2 (\frac{15}{2})^{2} - (35)^{2}$
Length $= 2 \frac{225}{4} - 45$

Final Computation

Length $= 2 \frac{225 - 180}{4}$
Length $= 2 \frac{45}{4} = 2 \cdot \frac{45}{2}$
Length $= 45 = 35$
The length of the chord is $35$ .

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are tasked with finding the length of a chord formed by a line that is tangent to the parabola

y^{2} = 30 x

and passes through the point

P (- 30, 0)

, while intersecting the circle

4 x^{2} + 4 y^{2} + 120 x + 675 = 0

Unveiling the Circle

First, we simplify the circle's equation by dividing by

4

x^{2} + y^{2} + 30 x + \frac{675}{4} = 0

Comparing this to the general form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, we identify the center

C

(- 15, 0)

. The radius

R

is calculated as follows:

R = (- 15)^{2} + 0^{2} - \frac{675}{4} = 225 - \frac{675}{4} = \frac{900 - 675}{4} = \frac{225}{4} = \frac{15}{2}

The Parabola's Secret

For the parabola

y^{2} = 30 x

, we identify

4 a = 30

, which gives

a = \frac{15}{2}

. The equation of a tangent line with slope

m

y = m x + \frac{a}{m}

Substituting

a = \frac{15}{2}

and the point

P (- 30, 0)

into the tangent equation:

0 = m (- 30) + \frac{15}{2 m} \Rightarrow 30 m = \frac{15}{2 m} \Rightarrow 60 m^{2} = 15 \Rightarrow m^{2} = \frac{1}{4}

Taking

m = \frac{1}{2}

, the equation of the line becomes

y = \frac{1}{2} x + 15

, which rearranges to:

x - 2 y + 30 = 0

The Grand Finale

The length of the chord

L

is given by

L = 2 R^{2} - p^{2}

, where

p

is the perpendicular distance from the center

C (- 15, 0)

to the line

x - 2 y + 30 = 0

Calculating

p

p = \frac{∣1 ( - 15 ) - 2 ( 0 ) + 30∣}{1 ^{2} + ( - 2 ) ^{2}} = \frac{∣ - 15 + 30∣}{5} = \frac{15}{5} = 35

Finally, we compute the chord length

L

L = 2 (\frac{15}{2})^{2} - (35)^{2} = 2 \frac{225}{4} - 45 = 2 \frac{225 - 180}{4} = 2 \frac{45}{4} = 45

The final length of the chord is $35$ .

Common Traps & FAQs

Wait, why do we get two slopes (

m = \pm 1/2

)? Does it matter which one I pick?

That is a fantastic observation! The point

P (- 30, 0)

lies outside the parabola

y^{2} = 30 x

. From any point outside a parabola, you can draw exactly two tangents. Both tangents will intersect the circle and form chords. However, due to the symmetry of the circle and the parabola about the x-axis, both chords will have the same length. So, picking either

m = 1/2

m = - 1/2

will lead you to the same final result.

Can I use the discriminant method to find the tangent instead of the

y = m x + a / m

formula?

Absolutely! The discriminant method is a powerful, fundamental tool. If you substitute

y = m x + c

into

y^{2} = 30 x

, you get a quadratic in

x

(m x + c)^{2} = 30 x

, or

m^{2} x^{2} + (2 m c - 30) x + c^{2} = 0

. For tangency, the discriminant

D

must be zero. Setting

D = (2 m c - 30)^{2} - 4 (m^{2}) (c^{2}) = 0

will yield the same condition

c = a / m

. It is a great way to verify your work!

Wait, the chord length formula

2 R^{2} - p^{2}

... is it always valid?

Yes, it is! This formula is derived directly from the Pythagorean theorem in the triangle formed by the circle's center, the midpoint of the chord, and one of the endpoints of the chord on the circle. As long as the line intersects the circle (i.e.,

p < R

), this formula will always give you the correct length of the chord.

Animated Solution for Mathematics - Conic Sections: If a line along a chord of the circle $4 x^{2} + 4 y^{2} + 120 x + 675 = 0$ , passes through the point $(- 30, 0)$ and is tangent to the parabola $y^{2} = 30 x$ , then the length of this chord is :

Select Answer:

Visualized Solution

Visualizing the Geometry

Analyzing the Circle

Calculating the Radius

Tangent to the Parabola

Finding the Slope $m$

Equation of the Line

The Chord and Perpendicular Distance

Calculating Perpendicular Distance $p$

Calculating Chord Length

Final Computation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Unveiling the Circle

The Parabola's Secret

The Grand Finale

Similar Questions

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 $tan θ =$

Let $A B$ be a chord of length 12 of the circle $(x - 2)^{2} + (y + 1)^{2} = \frac{169}{4}$ . If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$ , then five times the distance of point $P$ from chord $A B$ is equal to ________

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

Let $P$ be the point on the parabola $y^{2} = 4 x$ which is at the shortest distance from the center $S$ of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0$ . Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then

Let the shortest distance from $(a, 0)$ , $a > 0$ to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is:

Let a line perpendicular to the line $2 x - y = 10$ touch the parabola $y^{2} = 4 (x - 9)$ at the point $P$ . The distance of the point $P$ from the centre of the circle $x^{2} + y^{2} - 14 x - 8 y + 56 = 0$ is __________

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , is:

If two tangents drawn from a point $(α, β)$ lying on the ellipse $25 x^{2} + 4 y^{2} = 1$ to the parabola $y^{2} = 4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 α + 5)^{2} + (16 β^{2} + 50)^{2}$ equals ____.

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ is :-

The tangent to the parabola $y^{2} = 4 x$ at the point where it intersects the circle $x^{2} + y^{2} = 5$ in the first quadrant, passes through the point :

Select Answer:

Visualized Solution

Visualizing the Geometry

Analyzing the Circle

Calculating the Radius

Tangent to the Parabola

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

Equation of the Line

The Chord and Perpendicular Distance

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

Calculating Chord Length

Final Computation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Unveiling the Circle

The Parabola's Secret

The Grand Finale

Similar Questions

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 common tangent to the parabolas, y2=4x and x2=4y also touches the circle, x2+y2=c2, then c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola, y2=4x, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the circle C1​:x2+y2−2x−1=0 at the point (2, 1) cuts off a chord of length 4 from a circle C2​ whose centre is (3, -2). The radius of C2​ is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the parabola y2=4x at the point where it intersects the circle x2+y2=5 in the first quadrant, passes through the point :

Finding the Slope $m$

Calculating Perpendicular Distance $p$

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 common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , is:

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ is :-

The tangent to the parabola $y^{2} = 4 x$ at the point where it intersects the circle $x^{2} + y^{2} = 5$ in the first quadrant, passes through the point :