JEE Main 2019 (10 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: The tangent and normal to the ellipse $3 x^{2} + 5 y^{2} = 32$ at the point $P (2, 2)$ meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is :

Q: Why do we use implicit differentiation instead of just memorizing the tangent formula for an ellipse?

While formulas like $T=0$ are powerful, implicit differentiation is a fundamental tool that works for any curve, not just standard conics. In JEE Advanced, understanding the 'why' behind the slope is what separates a calculator from a physicist. Mastering the derivative ensures you can handle non-standard curves where memorized formulas fail.

Q: Why is the height of the triangle simply the y-coordinate of point P?

Because the base of the triangle lies on the x-axis (where $y=0$). The height of any triangle is the perpendicular distance from the vertex to the line containing the base. Since the base is on the x-axis, the vertical distance from $P(2, 2)$ to the x-axis is just the absolute value of its y-coordinate, which is 2.

Select Answer:

Progressive Hints

Hint 1

Visualize the geometry: You are dealing with a triangle where the base lies entirely on the x-axis. The vertices are the point of contact

P

and the two x-intercepts

Q

and

R

. Once you identify the base as the segment

QR

and the height as the perpendicular distance from

P

to the x-axis, the problem simplifies significantly.

Hint 2

You need the slopes of the tangent and normal lines. Use implicit differentiation on the ellipse equation

3 x^{2} + 5 y^{2} = 32

to find the general slope

\frac{d y}{d x}

. Remember, the normal is perpendicular to the tangent, so their slopes satisfy

m_{T} \cdot m_{N} = - 1

Hint 3

Start by calculating the derivative:

\frac{d y}{d x} = - \frac{3 x}{5 y}

. At point

P (2, 2)

, the tangent slope is

m_{T} = - \frac{3}{5}

. Use the point-slope form

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

to find the equations of both lines, then set

y = 0

to find the x-intercepts

Q

and

R

Visualized Solution

Visualizing the Ellipse and Point $P$

Given Ellipse: $3 x^{2} + 5 y^{2} = 32$
Point $P (2, 2)$
Verify $P$ on Ellipse: $3 (2)^{2} + 5 (2)^{2} = 12 + 20 = 32$ (True)

Differentiating to Find the Slope

Differentiate $3 x^{2} + 5 y^{2} = 32$ with respect to $x$ :
$\frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (5 y^{2}) = \frac{d}{d x} (32)$
$6 x + 10 y \frac{d y}{d x} = 0$

Isolating the Derivative $\frac{d y}{d x}$

Rearrange to solve for $\frac{d y}{d x}$ :
$10 y \frac{d y}{d x} = - 6 x$
$\frac{d y}{d x} = - \frac{6 x}{10 y} = - \frac{3 x}{5 y}$

Calculating Slope at $P (2, 2)$

At $P (2, 2)$ , substitute $x = 2, y = 2$ into $\frac{d y}{d x}$ :
$m_{T} = - \frac{3 ( 2 )}{5 ( 2 )} = - \frac{6}{10}$
$m_{T} = - \frac{3}{5}$

Equation of the Tangent Line

Using point-slope form: $y - y_{1} = m_{T} (x - x_{1})$
$y - 2 = - \frac{3}{5} (x - 2)$
$5 (y - 2) = - 3 (x - 2) ⟹ 5 y - 10 = - 3 x + 6$
Tangent Equation: $3 x + 5 y = 16$

Finding Point $Q$ on the x-axis

Point $Q$ is the x-intercept of the tangent ( $y = 0$ ):
$3 x + 5 (0) = 16 ⟹ 3 x = 16$
$x = \frac{16}{3}$
Point $Q (\frac{16}{3}, 0)$

Slope of the Normal Line

Normal is perpendicular to the tangent.
Slope of Normal $m_{N} = - \frac{1}{m _{T}}$
$m_{N} = - \frac{1}{- \frac{3}{5}} = \frac{5}{3}$

Equation of the Normal Line

Using point-slope form for the normal at $P (2, 2)$ :
$y - 2 = \frac{5}{3} (x - 2)$
$3 (y - 2) = 5 (x - 2) ⟹ 3 y - 6 = 5 x - 10$
Normal Equation: $5 x - 3 y = 4$

Finding Point $R$ on the x-axis

Point $R$ is the x-intercept of the normal ( $y = 0$ ):
$5 x - 3 (0) = 4 ⟹ 5 x = 4$
$x = \frac{4}{5}$
Point $R (\frac{4}{5}, 0)$

Visualizing Triangle $P QR$

Triangle $P QR$ vertices: $P (2, 2)$ , $Q (\frac{16}{3}, 0)$ , $R (\frac{4}{5}, 0)$
Base $QR$ lies on the x-axis.
Height of triangle = y-coordinate of $P = 2$

Calculating the Base Length $QR$

Base length $QR = ∣ x_{Q} - x_{R} ∣$
$QR = ∣ \frac{16}{3} - \frac{4}{5} ∣$
$QR = ∣ \frac{16 \times 5 - 4 \times 3}{15} ∣ = ∣ \frac{80 - 12}{15} ∣$
$QR = \frac{68}{15}$

Applying the Area Formula

Area of $△ P QR = \frac{1}{2} \times Base \times Height$
Area $= \frac{1}{2} \times \frac{68}{15} \times 2$
Area $= \frac{68}{15}$ sq. units

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are exploring the geometry of the ellipse defined by the equation:

3 x^{2} + 5 y^{2} = 32

Our objective is to determine the area of the triangle formed by the tangent and normal lines at point

P (2, 2)

as they intersect the

x

-axis.

Phase 1

The Calculus of Curves

To find the slope of the tangent at

P (2, 2)

, we perform implicit differentiation on the ellipse equation with respect to

x

\frac{d}{d x} (3 x^{2} + 5 y^{2}) = \frac{d}{d x} (32)

6 x + 10 y \frac{d y}{d x} = 0

Rearranging for the derivative, we obtain the general slope formula:

\frac{d y}{d x} = - \frac{3 x}{5 y}

Evaluating this at point

P (2, 2)

, the slope of the tangent

m_{T}

is:

m_{T} = - \frac{3 ( 2 )}{5 ( 2 )} = - \frac{3}{5}

Phase 2

Constructing the Lines

Using the point-slope form

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

for the tangent line at

P (2, 2)

with

m_{T} = - \frac{3}{5}

y - 2 = - \frac{3}{5} (x - 2)

5 y - 10 = - 3 x + 6 \Rightarrow 3 x + 5 y = 16

The normal line is perpendicular to the tangent, so its slope

m_{N}

is the negative reciprocal of

m_{T}

m_{N} = \frac{5}{3}

Applying the point-slope form for the normal line:

y - 2 = \frac{5}{3} (x - 2)

3 y - 6 = 5 x - 10 \Rightarrow 5 x - 3 y = 4

Phase 3

The Intersection and the Area

To find the

x

-intercepts

Q

and

R

, we set

y = 0

in both line equations. For the tangent line

3 x + 5 y = 16

3 x = 16 \Rightarrow x = \frac{16}{3}

Thus, point

Q

(\frac{16}{3}, 0)

. For the normal line

5 x - 3 y = 4

5 x = 4 \Rightarrow x = \frac{4}{5}

Thus, point

R

(\frac{4}{5}, 0)

The base of the triangle

QR

lies on the

x

-axis, with length:

Base = \frac{16}{3} - \frac{4}{5} = \frac{80 - 12}{15} = \frac{68}{15}

The height of the triangle is the vertical distance from

P (2, 2)

to the

x

-axis, which is

2

. The area is calculated as:

Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times \frac{68}{15} \times 2

The final area of the triangle is $\frac{68}{15}$ square units.

Common Traps & FAQs

Why do we use implicit differentiation instead of just memorizing the tangent formula for an ellipse?

While formulas like

T = 0

are powerful, implicit differentiation is a fundamental tool that works for any curve, not just standard conics. In JEE Advanced, understanding the 'why' behind the slope is what separates a calculator from a physicist. Mastering the derivative ensures you can handle non-standard curves where memorized formulas fail.

Why is the height of the triangle simply the y-coordinate of point P?

Because the base of the triangle lies on the x-axis (where

y = 0

). The height of any triangle is the perpendicular distance from the vertex to the line containing the base. Since the base is on the x-axis, the vertical distance from

P (2, 2)

to the x-axis is just the absolute value of its y-coordinate, which is 2.

Is there a faster way to find the x-intercepts without writing the full line equations?

Yes, you can use the intercept form of a line, but writing the full equation is safer and less prone to algebraic errors. Don't rush the algebra; the beauty of this problem lies in the precision of the final fraction. Taking the extra thirty seconds to write the full equation prevents silly mistakes that cost ranks.

Animated Solution for Mathematics - Conic Sections: The tangent and normal to the ellipse $3 x^{2} + 5 y^{2} = 32$ at the point $P (2, 2)$ meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is :

Select Answer:

Visualized Solution

Visualizing the Ellipse and Point $P$

Differentiating to Find the Slope

Isolating the Derivative $\frac{d y}{d x}$

Calculating Slope at $P (2, 2)$

Equation of the Tangent Line

Finding Point $Q$ on the x-axis

Slope of the Normal Line

Equation of the Normal Line

Finding Point $R$ on the x-axis

Visualizing Triangle $P QR$

Calculating the Base Length $QR$

Applying the Area Formula

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

If the tangent at a point P on the parabola $y^{2} = 3 x$ is parallel to the line $x + 2 y = 1$ and the tangents at the points Q and R on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$ are perpendicular to the line $x - y = 2$ , then the area of the triangle PQR is:

Let the tangent to the parabola $S : y^{2} = 2 x$ at the point $P (2, 2)$ meet the $x$ -axis at $Q$ and normal at it meet the parabola $S$ at the point $R$ . Then the area (in sq. units) of the triangle $P QR$ is equal to:

If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point P on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , is

The tangent and the normal lines at the point $(3, 1)$ to the circle $x^{2} + y^{2} = 4$ and the x-axis form a triangle. The area of this triangle (in square units) is :

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^{2} + y^{2} = 4$ at $(1, rac{\sqrt{3}}{1})$ is .........

Select Answer:

Visualized Solution

Visualizing the Ellipse and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Differentiating to Find the Slope

Isolating the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Calculating Slope at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2,2)

Equation of the Tangent Line

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

Slope of the Normal Line

Equation of the Normal Line

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

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

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

Applying the Area Formula

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normal to the ellipse 3x2+4y2=12 at a point P on it is parallel to the line, 2x+y=4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse 9x2​+5y2​=1, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent and the normal lines at the point (3​,1) to the circle x2+y2=4 and the x-axis form a triangle. The area of this triangle (in square units) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn at the point O(0,0) and P(1+5​,2) on the circle x2+y2−2x−4y=0 intersect at the point Q, then the area of the triangle OPQ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle x2+y2=4 at (1, rac{\sqrt{3}}{1}) is .........

Visualizing the Ellipse and Point $P$

Isolating the Derivative $\frac{d y}{d x}$

Calculating Slope at $P (2, 2)$

Finding Point $Q$ on the x-axis

Finding Point $R$ on the x-axis

Visualizing Triangle $P QR$

Calculating the Base Length $QR$

If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point P on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , is

The tangent and the normal lines at the point $(3, 1)$ to the circle $x^{2} + y^{2} = 4$ and the x-axis form a triangle. The area of this triangle (in square units) is :

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^{2} + y^{2} = 4$ at $(1, rac{\sqrt{3}}{1})$ is .........