Let a focus of the ellipse E: x^2/a^2 + y^2/b^2 = 1 be S(4, 0) and its eccentricity be 4/5. If the point P(3, α) lies on E and O is the origin, then the area of ΔPOS is equal to:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q9 – JEE-MAIN 2026

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Home Maths Conic SectionsQuestion #115

Let a focus of the ellipse E:

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

S (4, 0)

and its eccentricity be

\frac{4}{5}

. If the point

P (3, α)

lies on E and O is the origin, then the area of

△ P O S

is equal to:

Clue 1 of 3: Understand Ellipse Properties

Recall the standard equation of an ellipse and the relationship between its semi-major axis, semi-minor axis, and eccentricity.

🧠 Full Solution Path

Step 1: Determine the parameters of the ellipse✦ Active

Given a focus $S (4, 0)$ and eccentricity $e = \frac{4}{5}$ . For an ellipse with foci on the x-axis, the focus is $(a e, 0)$ . Thus, $a e = 4$ . Substituting $e = \frac{4}{5}$ , we get $a (\frac{4}{5}) = 4$ , which implies $a = 5$ . The relationship between $a$ , $b$ , and $e$ for an ellipse is $b^{2} = a^{2} (1 - e^{2})$ . Substituting the values of $a$ and $e$ :

b^{2} = 5^{2} (1 - {(\frac{4}{5})}^{2}) = 25 (1 - \frac{16}{25}) = 25 (\frac{9}{25}) = 9

So, $b = 3$ . The equation of the ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ .

Step 2: Find the y-coordinate of point P○ Expand

The point $P (3, α)$ lies on the ellipse. Substitute these coordinates into the ellipse equation:

\frac{3^{2}}{25} + \frac{α^{2}}{9} = 1 ⟹ \frac{9}{25} + \frac{α^{2}}{9} = 1

Solving for $α^{2}$ :

\frac{α^{2}}{9} = 1 - \frac{9}{25} = \frac{16}{25} ⟹ α^{2} = \frac{16 \times 9}{25} = \frac{144}{25}

Thus, $α = \pm \frac{12}{5}$ . We can choose $α = \frac{12}{5}$ for calculating the area, as the absolute value will be used.

Step 3: Calculate the area of

△ P O S

○ Expand

The vertices of the triangle are $O (0, 0)$ , $P (3, \frac{12}{5})$ , and $S (4, 0)$ . The area of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ is given by $\frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$ . Using $O (0, 0)$ , $P (3, \frac{12}{5})$ , $S (4, 0)$ :

Area = \frac{1}{2} | 0 (\frac{12}{5} - 0) + 3 (0 - 0) + 4 (0 - \frac{12}{5}) |

Simplifying the expression:

Area = \frac{1}{2} | 0 + 0 - \frac{48}{5} | = \frac{1}{2} \times \frac{48}{5} = \frac{24}{5}

💡 Teacher's Secret Hint

Remember that the area of a triangle must be positive. The choice of positive or negative $α$ does not affect the final area.

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Maths Question 9 – JEE-MAIN 2026

Clue 1 of 3: Understand Ellipse Properties

🧠 Full Solution Path