JEE Main 2021 (25 February Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: A tangent is drawn to the parabola $y^{2} = 6 x$ which is perpendicular to the line $2 x + y = 1$ . Which of the following points does NOT lie on it?

Q: Wait, why do we use $y = mx + \frac{a}{m}$ instead of the point-slope form?

Great question! We use $y = mx + \frac{a}{m}$ because it is the 'condition of tangency' specifically derived for a parabola. It automatically ensures that the line touches the curve at exactly one point. Using point-slope form would require you to know the point of contact first, which we don't have yet!

Q: How did you get $a = \frac{3}{2}$ so quickly?

Always compare your given equation to the standard form $y^2 = 4ax$. Here, $4a = 6$. Dividing both sides by 4 gives $a = \frac{6}{4}$, which simplifies beautifully to $\frac{3}{2}$. Never rush this step; a small error here cascades through the whole problem!

Q: Is there a way to solve this without the formula?

Yes! You could substitute $y = mx + c$ into $y^2 = 6x$, resulting in a quadratic in $x$. Since it is a tangent, the discriminant ($D = b^2 - 4ac$) must be zero. However, memorizing the standard tangent form is much faster for JEE Advanced.

Q: How do I know which point doesn't lie on the line?

Simply plug the $x$ and $y$ coordinates of each option into your derived equation $x - 2y + 6 = 0$. If the result is not zero, that point does not satisfy the equation and therefore does not lie on the line.

Select Answer:

Visualized Solution

Visualizing the Parabola $y^{2} = 6 x$

Parabola: $y^{2} = 6 x$
Standard form: $y^{2} = 4 a x$

The Given Line $2 x + y = 1$

Given Line: $2 x + y = 1$
The tangent must be perpendicular to this line.

Slope of the Given Line

Equation: $2 x + y = 1$
Slope-intercept form: $y = - 2 x + 1$
Slope ( $m_{1}$ ) = $- 2$

Perpendicularity Condition

Condition for perpendicular lines: $m_{1} \times m_{2} = - 1$

Finding Tangent Slope $m$

$- 2 \times m = - 1$
Slope of tangent ( $m$ ) = $\frac{1}{2}$

Parabola Parameter $a$

Compare $y^{2} = 6 x$ with $y^{2} = 4 a x$
Coefficient of $x$ : $4 a = 6$

Calculating $a = \frac{3}{2}$

$a = \frac{6}{4}$
Simplified: $a = \frac{3}{2}$

Tangent Equation Formula

Equation of tangent with slope $m$ :
$y = m x + \frac{a}{m}$

Substituting $m$ and $a$

Substitute $m = \frac{1}{2}$ and $a = \frac{3}{2}$ :
$y = \frac{1}{2} x + \frac{\frac{3}{2}}{\frac{1}{2}}$

Simplifying the Equation

$y = \frac{x}{2} + 3$
Multiply by $2$ : $2 y = x + 6$
Standard form: $x - 2 y + 6 = 0$

Checking Point $(5, 4)$

Check point $(5, 4)$ in $x - 2 y + 6 = 0$ :
LHS: $5 - 2 (4) + 6$
$= 5 - 8 + 6 = 3$
Since $3 \neq = 0$ , $(5, 4)$ is NOT on the line.

Final Conclusion

Point $(5, 4)$ does NOT lie on the tangent.
The equation of the tangent is $x - 2 y + 6 = 0$ .
Correct Option: (5, 4)

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given parabola is defined by the equation

y^{2} = 6 x

. By comparing this to the standard form

y^{2} = 4 a x

, we identify the parameter

a

4 a = 6 ⟹ a = \frac{6}{4} = \frac{3}{2}

This value

a = \frac{3}{2}

represents the fundamental geometric constant of our parabola.

Determining the Slope

We are given the line

2 x + y = 1

. Rewriting this in slope-intercept form (

y = m x + c

), we obtain

y = - 2 x + 1

. The slope of this line is

m_{1} = - 2

Since our tangent line must be perpendicular to this line, the product of their slopes must be

- 1

. Let

m

be the slope of the tangent:

m_{1} \times m = - 1 ⟹ - 2 \times m = - 1 ⟹ m = \frac{1}{2}

The Master Equation of Tangency

The equation of a line with slope

m

that is tangent to the parabola

y^{2} = 4 a x

is given by the formula:

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

Substituting our known values

m = \frac{1}{2}

and

a = \frac{3}{2}

into this formula, we get:

y = (\frac{1}{2}) x + \frac{\frac{3}{2}}{\frac{1}{2}}

Final Calculation

Simplifying the constant term, we find that

\frac{3/2}{1/2} = 3

. Thus, the equation of the tangent line is:

y = \frac{x}{2} + 3

Multiplying the entire equation by

2

to clear the fraction, we obtain

2 y = x + 6

. Rearranging this into the standard linear form, we arrive at the final equation of the tangent:

x - 2 y + 6 = 0

Verification of Points

To determine if a point lies on this tangent, we substitute its coordinates into the equation

x - 2 y + 6 = 0

. For the point

(5, 4)

L H S = 5 - 2 (4) + 6 = 5 - 8 + 6 = 3

Since $3 eq 0$, the point

(5, 4)

does not satisfy the equation. Therefore, the point

(5, 4)

does not lie on the tangent line.

Common Traps & FAQs

Wait, why do we use

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

instead of the point-slope form?

Great question! We use

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

because it is the 'condition of tangency' specifically derived for a parabola. It automatically ensures that the line touches the curve at exactly one point. Using point-slope form would require you to know the point of contact first, which we don't have yet!

How did you get

a = \frac{3}{2}

so quickly?

Always compare your given equation to the standard form

y^{2} = 4 a x

. Here,

4 a = 6

. Dividing both sides by 4 gives

a = \frac{6}{4}

, which simplifies beautifully to

\frac{3}{2}

. Never rush this step; a small error here cascades through the whole problem!

Why is the slope of the tangent

\frac{1}{2}

and not

- 2

Remember the rule for perpendicular lines: the product of their slopes must be

- 1

. If the given line

2 x + y = 1

has a slope of

- 2

, then

- 2 \times m = - 1

forces

m

to be

\frac{1}{2}

. Always check your signs!

Is there a way to solve this without the formula?

Yes! You could substitute

y = m x + c

into

y^{2} = 6 x

, resulting in a quadratic in

x

. Since it is a tangent, the discriminant (

D = b^{2} - 4 a c

) must be zero. However, memorizing the standard tangent form is much faster for JEE Advanced.

How do I know which point doesn't lie on the line?

Simply plug the

x

and

y

coordinates of each option into your derived equation

x - 2 y + 6 = 0

. If the result is not zero, that point does not satisfy the equation and therefore does not lie on the line.

Animated Solution for Mathematics - Conic Sections: A tangent is drawn to the parabola $y^{2} = 6 x$ which is perpendicular to the line $2 x + y = 1$ . Which of the following points does NOT lie on it?

Select Answer:

Visualized Solution

Visualizing the Parabola $y^{2} = 6 x$

The Given Line $2 x + y = 1$

Slope of the Given Line

Perpendicularity Condition

Finding Tangent Slope $m$

Parabola Parameter $a$

Calculating $a = \frac{3}{2}$

Tangent Equation Formula

Substituting $m$ and $a$

Simplifying the Equation

Checking Point $(5, 4)$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Determining the Slope

The Master Equation of Tangency

Final Calculation

Verification of Points

Similar Questions

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 :

The equation of a tangent to the parabola $y^{2} = 8 x$ is $y = x + 2$ . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

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

Let the tangent drawn to the parabola $y^{2} = 24 x$ at the point $(α, β)$ is perpendicular to the line $2 x + 2 y = 5$ . Then the normal to the hyperbola $\frac{x ^{2}}{α ^{2}} - \frac{y ^{2}}{β ^{2}} = 1$ at the point $(α + 4, β + 4)$ does NOT pass through the point :

Tangent to the curve $y = x^{2} + 6$ at a point $(1, 7)$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are

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 :

The point of intersection of the tangents at the ends of the latus rectum of the parabola $y^{2} = 4 x$ is.........

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

Let the normal at the point $P$ on the parabola $y^{2} = 6 x$ pass through the point $(5, - 8)$ . If the tangent at $P$ to the parabola intersects its directrix at the point $Q$ , then the ordinate of the point $Q$ is :

If a circle touches y-axis at $(0, 4)$ and passes through $(2, 0)$ then which of the following can-not be the tangent to the circle

Select Answer:

Visualized Solution

Visualizing the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=6x

The Given Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x+y=1

Slope of the Given Line

Perpendicularity Condition

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

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=23​

Tangent Equation Formula

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

Simplifying the Equation

Checking Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (5,4)

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Determining the Slope

The Master Equation of Tangency

Final Calculation

Verification of Points

Similar Questions

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a tangent to the parabola y2=8x is y=x+2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

.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; } Tangent to the curve y=x2+6 at a point (1,7) touches the circle x2+y2+16x+12y+c=0 at a point Q. Then the coordinates of Q are

.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; } The point of intersection of the tangents at the ends of the latus rectum of the parabola y2=4x is.........

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A possible equation of L is

A common tangent of the two circles is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a circle touches y-axis at (0,4) and passes through (2,0) then which of the following can-not be the tangent to the circle

Visualizing the Parabola $y^{2} = 6 x$

The Given Line $2 x + y = 1$

Finding Tangent Slope $m$

Parabola Parameter $a$

Calculating $a = \frac{3}{2}$

Substituting $m$ and $a$

Checking Point $(5, 4)$

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 :

The equation of a tangent to the parabola $y^{2} = 8 x$ is $y = x + 2$ . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

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

Tangent to the curve $y = x^{2} + 6$ at a point $(1, 7)$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are

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 :

The point of intersection of the tangents at the ends of the latus rectum of the parabola $y^{2} = 4 x$ is.........

A possible equation of $L$ is

If a circle touches y-axis at $(0, 4)$ and passes through $(2, 0)$ then which of the following can-not be the tangent to the circle