Problem 161 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The tangent and normal to the ellipse \(3 x^{2}+5 y^{2}=32\) at the point \(\mathrm{P}(2,2)\) meet the \(\mathrm{x}\)-axis at \(\mathrm{Q}\) and \(\mathrm{R}\), respectively. Then the area (in sq. units) of the triangle \(\mathrm{PQR}\) is: (a) \(\frac{34}{15}\) (b) \(\frac{14}{3}\) (c) \(\frac{16}{3}\) (d) \(\frac{68}{15}\)

Step-by-Step Solution

Answer

The area of triangle PQR is \(\frac{68}{15}\) square units (option d).

Step 1: Equation of Tangent at Point P

The equation of the tangent to the ellipse \(3x^2 + 5y^2 = 32\) at point \((2, 2)\) is given by: \(3 \cdot 2x + 5 \cdot 2y = 32\). Simplifying, we get \(6x + 10y = 32\).

Step 2: Find Point Q (Intersection of Tangent and x-axis)

To find where the tangent meets the \(x\)-axis (point \(Q\)), set \(y=0\) in the tangent equation from Step 1: \(6x + 10 \cdot 0 = 32\). Solving for \(x\), \(x = \frac{32}{6} = \frac{16}{3}\). Hence, the coordinates of point \(Q\) are \(\left(\frac{16}{3}, 0\right)\).

Step 3: Equation of Normal at Point P

The slope of the tangent at \((2, 2)\) is \(-\frac{6}{10} = -\frac{3}{5}\). Therefore, the slope of the normal is \(\frac{5}{3}\). Using point-slope form for the normal line from point \((2, 2)\): \(y - 2 = \frac{5}{3}(x - 2)\). This simplifies to \(3y - 6 = 5x - 10\), or \(5x - 3y = 4\).

Step 4: Find Point R (Intersection of Normal and x-axis)

To find where the normal meets the \(x\)-axis (point \(R\)), set \(y=0\) in the normal equation from Step 3: \(5x - 3 \cdot 0 = 4\). Solving for \(x\), \(x = \frac{4}{5}\). Hence, the coordinates of point \(R\) are \(\left(\frac{4}{5}, 0\right)\).

Step 5: Calculate Area of Triangle PQR

The points of the triangle are \(P(2, 2)\), \(Q\left(\frac{16}{3}, 0\right)\), and \(R\left(\frac{4}{5}, 0\right)\). The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by \[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\]Substituting the given points, we have:\[\text{Area} = \frac{1}{2} \left| 2(0-0) + \frac{16}{3}(0-2) + \frac{4}{5}(2-0) \right|\]\[= \frac{1}{2} \left| -\frac{32}{3} + \frac{8}{5} \right|\]Finding a common denominator (15):\(-\frac{160}{15} + \frac{24}{15} = -\frac{136}{15}\)\[\text{Area} = \frac{1}{2} \times \frac{136}{15} = \frac{68}{15}\]Thus, the area of the triangle \(PQR\) is \(\frac{68}{15}\).

Step 6: Conclusion

The solution reveals that the area of triangle \(PQR\) formed by the tangent and normal to the ellipse with the \(x\)-axis is option (d): \(\frac{68}{15}\).

Key Concepts

Equation of Tangent

When finding the equation of a tangent to an ellipse, it helps to start with the equation of the ellipse itself. In this exercise, the ellipse is defined by the equation \(3x^2 + 5y^2 = 32\). A tangent line is a straight line that touches the ellipse at exactly one point, without crossing it.
To find the tangent line at a specific point, like \((2, 2)\) in this case, we use a derivative approach or the standard formula. The equation of the tangent at any point \((x_1, y_1)\) on an ellipse \(Ax^2 + By^2 = C\) is given by \(Ax_1x + By_1y = C\). This formula comes from the fact that the tangent line represents the slope of a curve at a single point.
In the exercise, substituting the values for the point \((2, 2)\), we simplify the equation:

• Given: \(3 \cdot 2x + 5 \cdot 2y = 32\)
• Turns into: \(6x + 10y = 32\)

Thus, the equation \(6x + 10y = 32\) is the equation of the tangent line that touches the ellipse at the point \((2, 2)\).

Equation of Normal

The concept of a normal line is closely related to that of a tangent. Whereas a tangent line touches a curve at one point, a normal line is perpendicular (or at a right angle) to the tangent line at the point of tangency.
The slope of the tangent line in this exercise was found to be \(-\frac{3}{5}\). This is calculated by arranging the tangent line equation \(6x + 10y = 32\) into slope-intercept form \(y = mx + c\), where \(m\) is the slope.
The perpendicular (normal) slope is the negative reciprocal of the tangent's slope. Therefore, if the tangent slope is \(-\frac{3}{5}\), then the slope of the normal is \(\frac{5}{3}\).

• Using point-slope formula for lines: \(y - y_1 = m(x - x_1)\), for point \((2, 2)\) and slope \(\frac{5}{3}\)
• The equation becomes: \(y - 2 = \frac{5}{3}(x - 2)\)
• Which simplifies to: \(5x - 3y = 4\)

This equation describes the normal line at the point \((2, 2)\) on the ellipse.

Area of Triangle

Calculating the area of a triangle is essential when you're dealing with shapes formed by various lines such as tangents and normals to an ellipse. Especially when these lines intersect with the axes to form distinct vertices of the triangle.
In this problem, we find the area of triangle \(PQR\) where points \(Q\) and \(R\) are the intersection points of the tangent and normal with the \(x\)-axis, respectively, and \(P(2, 2)\) is the point of tangency.

• Point \(Q\) is calculated by setting \(y = 0\) in the tangent equation and solving for \(x\), giving \(Q\left(\frac{16}{3}, 0\right)\).
• Point \(R\) is found similarly, using the normal's equation, leading to \(R\left(\frac{4}{5}, 0\right)\).
• The area formula used is: \[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\]

By substituting these points:

• \(\frac{1}{2} \left( -\frac{32}{3} + \frac{8}{5} \right)\)
• This arithmetic allows us to compute the common denominator, yielding an area of \(\frac{68}{15}\).

Thus, the area of the triangle \(PQR\) is \(\frac{68}{15}\) square units.