JEE Main 2019 (12 April)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: The tangents to the curve $y = (x - 2)^{2} - 1$ at its points of intersection with the line $x - y = 3$ , intersect at the point :

Q: Why can't we just use the property that the intersection of tangents to a parabola at $x_1$ and $x_2$ is at $(\frac{x_1+x_2}{2}, \dots)$?

That is a brilliant observation! For a standard parabola $y = ax^2$, the intersection of tangents at $x_1$ and $x_2$ does indeed occur at the $x$-coordinate $\frac{x_1+x_2}{2}$. However, our parabola is shifted: $y = (x - 2)^2 - 1$. You can still use the property, but you must be careful with the transformation. If you shift the parabola back to the origin, the property holds, and then you shift the result back. Calculating it directly as we did is often safer to avoid coordinate-shift errors!

Q: Is the slope really zero at $x = 2$?

Yes, absolutely! Look at the equation $y = (x - 2)^2 - 1$. When $x = 2$, the term $(x - 2)^2$ becomes zero, which is the minimum value of the parabola (the vertex). At the vertex of a parabola, the tangent is always horizontal, meaning its slope is zero. Our derivative $\frac{dy}{dx} = 2(x - 2)$ confirms this: at $x = 2$, the slope is $2(2 - 2) = 0$. It is a perfect, flat landing.

Select Answer:

Progressive Hints

Hint 1

Hint 1: To find the intersection points of the line

x - y = 3

and the parabola

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

, substitute

y = x - 3

into the parabola equation. This will give you a quadratic equation in

x

. Solve this to find the

x

-coordinates of the intersection points.

Hint 2

Hint 2: Once you have the intersection points, you need the slopes of the tangents at these points. Recall that the slope of the tangent to a curve

y = f (x)

at a point

(x_{0}, y_{0})

is given by the derivative

\frac{d y}{d x}

evaluated at

x = x_{0}

. Calculate the derivative of the parabola equation and evaluate it at your intersection

x

-values.

Hint 3

Hint 3: With the slopes and the points of contact, write the equations of the two tangent lines using the point-slope form:

y - y_{1} = m (x - x_{1})

. Finally, solve the system of two linear equations to find the intersection point of the tangents.

Visualized Solution

Visualizing the Curve $y = (x - 2)^{2} - 1$

Given curve: $y = (x - 2)^{2} - 1$
This is an upward-opening parabola with vertex at $(2, - 1)$ .

The Intersecting Line $x - y = 3$

Given line: $x - y = 3 \Rightarrow y = x - 3$
We need to find the points where this line intersects the parabola.

Setting up the Intersection Equation

Substitute $y = x - 3$ into $y = (x - 2)^{2} - 1$ :
$(x - 2)^{2} - 1 = x - 3$

Solving the Quadratic Equation

Expand: $x^{2} - 4 x + 4 - 1 = x - 3$
Simplify: $x^{2} - 4 x + 3 = x - 3$
Rearrange: $x^{2} - 5 x + 6 = 0$

Finding Intersection Points $A$ and $B$

Factorize: $(x - 2) (x - 3) = 0$
Roots: $x = 2, x = 3$
For $x = 2, y = 2 - 3 = - 1 \Rightarrow A (2, - 1)$
For $x = 3, y = 3 - 3 = 0 \Rightarrow B (3, 0)$

Differentiating to Find Slopes

Differentiate $y = (x - 2)^{2} - 1$ with respect to $x$ :
$\frac{d y}{d x} = 2 (x - 2)$
This gives the slope of the tangent at any point $(x, y)$ on the curve.

Slope and Tangent at Point $A (2, - 1)$

At $A (2, - 1)$ , slope $m_{1} = 2 (2 - 2) = 0$
Equation of tangent at $A$ : $y - (- 1) = 0 (x - 2)$
Tangent 1: $y = - 1$

Slope and Tangent at Point $B (3, 0)$

At $B (3, 0)$ , slope $m_{2} = 2 (3 - 2) = 2$
Equation of tangent at $B$ : $y - 0 = 2 (x - 3)$
Tangent 2: $y = 2 x - 6$

Solving for the Intersection of Tangents

Solve the system:
1) $y = - 1$
2) $y = 2 x - 6$
Substitute (1) into (2): $- 1 = 2 x - 6$

Final Result and Conclusion

Solve for $x$ : $2 x = 5 \Rightarrow x = \frac{5}{2}$
The tangents intersect at $(\frac{5}{2}, - 1)$ .

00:00 / 00:00

The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

The parabola is defined by the equation

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

, which has its vertex at

(2, - 1)

. We are intersecting this curve with the straight line

x - y = 3

, which can be rewritten as

y = x - 3

The Intersection

To find the points of intersection, we equate the two expressions for

y

(x - 2)^{2} - 1 = x - 3

Expanding the left side, we obtain:

x^{2} - 4 x + 4 - 1 = x - 3

x^{2} - 4 x + 3 = x - 3

Rearranging all terms to one side yields the quadratic equation:

x^{2} - 5 x + 6 = 0

Factoring the quadratic gives

(x - 2) (x - 3) = 0

, resulting in roots

x = 2

and

x = 3

. Substituting these into

y = x - 3

, we find the intersection points:

For

x = 2

y = - 1

, giving point

A (2, - 1)

For

x = 3

y = 0

, giving point

B (3, 0)

The Tangents

To find the tangents, we calculate the derivative of the parabola:

\frac{d y}{d x} = 2 (x - 2)

For point

A (2, - 1)

, the slope is

m_{1} = 2 (2 - 2) = 0

. The tangent is a horizontal line passing through

(2, - 1)

, so its equation is:

y = - 1

For point

B (3, 0)

, the slope is

m_{2} = 2 (3 - 2) = 2

. Using the point-slope form

y - y_{1} = m (x - x_{1})

, we get:

y - 0 = 2 (x - 3) \Rightarrow y = 2 x - 6

Final Calculation

We now find the intersection of the two tangent lines

y = - 1

and

y = 2 x - 6

. Substituting

y = - 1

into the second equation:

- 1 = 2 x - 6

5 = 2 x

x = \frac{5}{2}

The intersection point of the two tangents is $(\frac{5}{2}, - 1)$ .

Common Traps & FAQs

Wait, why do we reject the negative root here? Or did we even have one?

In this specific problem, the quadratic equation we derived,

x^{2} - 5 x + 6 = 0

, factors beautifully into

(x - 2) (x - 3) = 0

. Both roots,

x = 2

and

x = 3

, are valid and represent the two distinct points where the line slices through the parabola. We don't reject either because both are valid intersection points on the curve.

Why can't we just use the property that the intersection of tangents to a parabola at

x_{1}

and

x_{2}

is at

(\frac{x _{1} + x _{2}}{2}, \dots)

That is a brilliant observation! For a standard parabola

y = a x^{2}

, the intersection of tangents at

x_{1}

and

x_{2}

does indeed occur at the

x

-coordinate

\frac{x _{1} + x _{2}}{2}

. However, our parabola is shifted:

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

. You can still use the property, but you must be careful with the transformation. If you shift the parabola back to the origin, the property holds, and then you shift the result back. Calculating it directly as we did is often safer to avoid coordinate-shift errors!

Is the slope really zero at

x = 2

Yes, absolutely! Look at the equation

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

. When

x = 2

, the term

(x - 2)^{2}

becomes zero, which is the minimum value of the parabola (the vertex). At the vertex of a parabola, the tangent is always horizontal, meaning its slope is zero. Our derivative

\frac{d y}{d x} = 2 (x - 2)

confirms this: at

x = 2

, the slope is

2 (2 - 2) = 0

. It is a perfect, flat landing.

Animated Solution for Mathematics - Differentiation: The tangents to the curve $y = (x - 2)^{2} - 1$ at its points of intersection with the line $x - y = 3$ , intersect at the point :

Select Answer:

Visualized Solution

Visualizing the Curve $y = (x - 2)^{2} - 1$

The Intersecting Line $x - y = 3$

Setting up the Intersection Equation

Solving the Quadratic Equation

Finding Intersection Points $A$ and $B$

Differentiating to Find Slopes

Slope and Tangent at Point $A (2, - 1)$

Slope and Tangent at Point $B (3, 0)$

Solving for the Intersection of Tangents

Final Result and Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Intersection

The Tangents

Final Calculation

Similar Questions

The tangent to the curve $y = x^{2} - 5 x + 5$ , parallel to the line $2 y = 4 x + 1$ , also passes through the point.

Which of the following points lies on the tangent to the curve $x^{4} e^{y} + 2 y + 1 = 3$ at the point $(1, 0)$ ?

The tangent to the curve, $y = x e^{x^{2}}$ passing through the point $(1, e)$ also passes through the point :

If the tangent to the curve $y = \frac{x}{x ^{2} - 3}, x \in R, x \neq = \pm 3$ , at a point $(α, β) \neq = (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then :

The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point where the curve intersects the y-axis passes through the point:

Let $S$ be the set of all values of $x$ for which the tangent to the curve $y = f (x) = x^{3} - x^{2} - 2 x$ at $(x, y)$ is parallel to the line segment joining the points $(1, f (1))$ and $(- 1, f (- 1))$ , then $S$ is equal to :

If $∣ f (x_{1}) - f (x_{2}) ∣ < (x_{1} - x_{2})^{2}$ , for all $x_{1}, x_{2} \in R$ . Find the equation of tangent to the curve $y = f (x)$ at the point $(1, 2)$ .

Angle between the tangents to the curve $y = x^{2} - 5 x + 6$ at the points $(2, 0)$ and $(3, 0)$ is

If the lines $x + y = a$ and $x - y = b$ touch the curve $y = x^{2} - 3 x + 2$ at the points where the curve intersects the $x$ -axis, then $\frac{a}{b}$ is equal to

If the tangent to the curve $y = x^{3}$ at the point $P (t, t^{3})$ meets the curve again at $Q$ , then the ordinate of the point which divides $P Q$ internally in the ratio $1 : 2$ is :

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangents to the curve y=(x−2)2−1 at its points of intersection with the line x−y=3, intersect at the point :

Select Answer:

Visualized Solution

Visualizing the Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=(x−2)2−1

The Intersecting Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−y=3

Setting up the Intersection Equation

Solving the Quadratic Equation

Finding Intersection Points .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

Differentiating to Find Slopes

Slope and Tangent at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(2,−1)

Slope and Tangent at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B(3,0)

Solving for the Intersection of Tangents

Final Result and Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Intersection

The Tangents

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the curve y=x2−5x+5, parallel to the line 2y=4x+1, also passes through the point.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following points lies on the tangent to the curve x4ey+2y​+1=3 at the point (1,0)?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the curve, y=xex2 passing through the point (1,e) also passes through the point :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the curve y=x2−3x​,x∈R,x=±3​, at a point (α,β)=(0,0) on it is parallel to the line 2x+6y−11=0, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The normal to the curve y(x−2)(x−3)=x+6 at the point where the curve intersects the y-axis passes through the point:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∣f(x1​)−f(x2​)∣<(x1​−x2​)2, for all x1​,x2​∈R. Find the equation of tangent to the curve y=f(x) at the point (1,2).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle between the tangents to the curve y=x2−5x+6 at the points (2,0) and (3,0) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines x+y=a and x−y=b touch the curve y=x2−3x+2 at the points where the curve intersects the x-axis, then ba​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the curve y=x3 at the point P(t,t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is :

Animated Solution for Mathematics - Differentiation: The tangents to the curve $y = (x - 2)^{2} - 1$ at its points of intersection with the line $x - y = 3$ , intersect at the point :

Visualizing the Curve $y = (x - 2)^{2} - 1$

The Intersecting Line $x - y = 3$

Finding Intersection Points $A$ and $B$

Slope and Tangent at Point $A (2, - 1)$

Slope and Tangent at Point $B (3, 0)$

The tangent to the curve $y = x^{2} - 5 x + 5$ , parallel to the line $2 y = 4 x + 1$ , also passes through the point.

Which of the following points lies on the tangent to the curve $x^{4} e^{y} + 2 y + 1 = 3$ at the point $(1, 0)$ ?

The tangent to the curve, $y = x e^{x^{2}}$ passing through the point $(1, e)$ also passes through the point :

If the tangent to the curve $y = \frac{x}{x ^{2} - 3}, x \in R, x \neq = \pm 3$ , at a point $(α, β) \neq = (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then :

The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point where the curve intersects the y-axis passes through the point:

If $∣ f (x_{1}) - f (x_{2}) ∣ < (x_{1} - x_{2})^{2}$ , for all $x_{1}, x_{2} \in R$ . Find the equation of tangent to the curve $y = f (x)$ at the point $(1, 2)$ .

Angle between the tangents to the curve $y = x^{2} - 5 x + 6$ at the points $(2, 0)$ and $(3, 0)$ is

If the lines $x + y = a$ and $x - y = b$ touch the curve $y = x^{2} - 3 x + 2$ at the points where the curve intersects the $x$ -axis, then $\frac{a}{b}$ is equal to

If the tangent to the curve $y = x^{3}$ at the point $P (t, t^{3})$ meets the curve again at $Q$ , then the ordinate of the point which divides $P Q$ internally in the ratio $1 : 2$ is :