JEE MainMathematics

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Finding $a^{2} + b^{2}$ for an Ellipse Given Eccentricity and a Point

Q: Why is e^2 = 1 - b^2/a^2 and not 1 - a^2/b^2?

The stem specifies a > b, which means the major axis is along the x-axis, so b^2 = a^2(1 - e^2).

Let the eccentricity of an ellipse (x^2)/(a^2)+(y^2)/(b^2)=1, a>b, be 1/4. If this ellipse passes through the point (-4 √(2/5), 3), then a^2+b^2 is equal to:

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A29B31C32D34

Step-by-step solutionView

Correct answer

Using $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{1}{16}$ , we find $b^{2} = \frac{15}{16} a^{2}$ ; substituting $(- 4 \sqrt{\frac{2}{5}}, 3)$ into the ellipse equation yields $a^{2} = 16$ and $b^{2} = 15$ , so $a^{2} + b^{2} = 31$ .

Option analysis

Why each option works or fails

A · 29

The student miscalculated the relation between $a^{2}$ and $b^{2}$ or made an arithmetic error when evaluating $x^{2} = {(- 4 \sqrt{\frac{2}{5}})}^{2} = \frac{32}{5}$ , arriving at $a^{2} + b^{2} = 29$ . Ensure ${(- 4 \sqrt{\frac{2}{5}})}^{2} = 16 \times \frac{2}{5} = \frac{32}{5}$ and substitute $b^{2} = \frac{15}{16} a^{2}$ carefully to obtain $a^{2} = 16$ and $b^{2} = 15$ .

B · 31

This is the correct option obtained by setting $b^{2} = a^{2} (1 - e^{2}) = \frac{15}{16} a^{2}$ , solving $\frac{32 / 5}{a^{2}} + \frac{9}{(15 / 16) a^{2}} = 1$ to get $a^{2} = 16$ , $b^{2} = 15$ , and summing to $31$ .

C · 32

The student computed $2 a^{2}$ instead of $a^{2} + b^{2}$ , yielding $2 (16) = 32$ , answering an intermediate property rather than the requested sum. Read the required target carefully: the question asks for $a^{2} + b^{2}$ , which is $16 + 15 = 31$ , not $2 a^{2}$ .

D · 34

The student used the relation $b^{2} = a^{2} (1 - e) = a^{2} (1 - \frac{1}{4}) = \frac{3}{4} a^{2}$ instead of squaring the eccentricity, leading to an incorrect pair of values for $a^{2}$ and $b^{2}$ . Remember that for a standard ellipse with $a > b$ , the eccentricity relation involves $e^{2}$ , so $b^{2} = a^{2} (1 - e^{2})$ , not $a^{2} (1 - e)$ .

Reviewed route

Solution

StepWorking

01Given

Ellipse equation is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a > b$ , eccentricity $e = \frac{1}{4}$ , and it passes through $(- 4 \sqrt{\frac{2}{5}}, 3)$ .

02Goal

Find the value of $a^{2} + b^{2}$ .

03Approach

Express $b^{2}$ in terms of $a^{2}$ using the eccentricity relation $b^{2} = a^{2} (1 - e^{2})$ . Then substitute the given point into the ellipse equation to solve for $a^{2}$ , then $b^{2}$ , and finally compute $a^{2} + b^{2}$ .

04Execute

Using $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{1}{16}$ , we get $\frac{b^{2}}{a^{2}} = 1 - \frac{1}{16} = \frac{15}{16}$ , which gives $b^{2} = \frac{15}{16} a^{2}$ .

05Execute

Substitute the point $(- 4 \sqrt{\frac{2}{5}}, 3)$ into $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ : $\frac{16 \times \frac{2}{5}}{a^{2}} + \frac{9}{\frac{15}{16} a^{2}} = 1 ⟹ \frac{32}{5 a^{2}} + \frac{9 \times 16}{15 a^{2}} = 1$ $\frac{32}{5 a^{2}} + \frac{48}{5 a^{2}} = 1 ⟹ \frac{80}{5 a^{2}} = 1 ⟹ \frac{16}{a^{2}} = 1 ⟹ a^{2} = 16.$

06Execute

Calculate $b^{2} = \frac{15}{16} a^{2} = \frac{15}{16} \times 16 = 15$ . Thus, $a^{2} + b^{2} = 16 + 15 = 31$ .

✓Verify

Check: $a^{2} = 16, b^{2} = 15 ⟹ e = \sqrt{1 - 15 / 16} = 1 / 4$ . Point check: $\frac{32 / 5}{16} + \frac{9}{15} = \frac{2}{5} + \frac{3}{5} = 1$ . Both conditions satisfied exactly.

Hints that build this answer step by step

For an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a > b$ and eccentricity $e = \frac{1}{4}$ , what is the relationship between $b^{2}$ and $a^{2}$ ?
$b^{2} = a^{2} (1 - \frac{1}{16}) = \frac{15}{16} a^{2}$
Substitute the point $(- 4 \sqrt{\frac{2}{5}}, 3)$ and $b^{2} = \frac{15}{16} a^{2}$ into the ellipse equation. What is the value of $a^{2}$ ?
$a^{2} = 16$
Now that $a^{2} = 16$ , what is $a^{2} + b^{2}$ ?
$31$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why is

e^{2} = 1 - b^{2} / a^{2}

and not

1 - a^{2} / b^{2}

The stem specifies $a > b$ , which means the major axis is along the x-axis, so $b^{2} = a^{2} (1 - e^{2})$ .

Answer

Using $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{1}{16}$ , we find $b^{2} = \frac{15}{16} a^{2}$ ; substituting $(- 4 \sqrt{\frac{2}{5}}, 3)$ into the ellipse equation yields $a^{2} = 16$ and $b^{2} = 15$ , so $a^{2} + b^{2} = 31$ .

Why each option works or fails

A: 29 - The student miscalculated the relation between $a^{2}$ and $b^{2}$ or made an arithmetic error when evaluating $x^{2} = {(- 4 \sqrt{\frac{2}{5}})}^{2} = \frac{32}{5}$ , arriving at $a^{2} + b^{2} = 29$ . Ensure ${(- 4 \sqrt{\frac{2}{5}})}^{2} = 16 \times \frac{2}{5} = \frac{32}{5}$ and substitute $b^{2} = \frac{15}{16} a^{2}$ carefully to obtain $a^{2} = 16$ and $b^{2} = 15$ .
B · correct: 31 - This is the correct option obtained by setting $b^{2} = a^{2} (1 - e^{2}) = \frac{15}{16} a^{2}$ , solving $\frac{32 / 5}{a^{2}} + \frac{9}{(15 / 16) a^{2}} = 1$ to get $a^{2} = 16$ , $b^{2} = 15$ , and summing to $31$ .
C: 32 - The student computed $2 a^{2}$ instead of $a^{2} + b^{2}$ , yielding $2 (16) = 32$ , answering an intermediate property rather than the requested sum. Read the required target carefully: the question asks for $a^{2} + b^{2}$ , which is $16 + 15 = 31$ , not $2 a^{2}$ .
D: 34 - The student used the relation $b^{2} = a^{2} (1 - e) = a^{2} (1 - \frac{1}{4}) = \frac{3}{4} a^{2}$ instead of squaring the eccentricity, leading to an incorrect pair of values for $a^{2}$ and $b^{2}$ . Remember that for a standard ellipse with $b^{2} = 15$ , the eccentricity relation involves $e^{2}$ , so $b^{2} = a^{2} (1 - e^{2})$ , not $a^{2} (1 - e)$ .

Step-by-step solution

Given: Ellipse equation is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a > b$ , eccentricity $e = \frac{1}{4}$ , and it passes through $(- 4 \sqrt{\frac{2}{5}}, 3)$ .
Goal: Find the value of $a^{2} + b^{2}$ .
Approach: Express $b^{2}$ in terms of $a^{2}$ using the eccentricity relation $b^{2} = a^{2} (1 - e^{2})$ . Then substitute the given point into the ellipse equation to solve for $a^{2}$ , then $b^{2}$ , and finally compute $a^{2} + b^{2}$ .
Execute: Using $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{1}{16}$ , we get $e^{2}$ , which gives $b^{2} = \frac{15}{16} a^{2}$ .
Execute: Substitute the point $(- 4 \sqrt{\frac{2}{5}}, 3)$ into $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ : $\frac{16 \times \frac{2}{5}}{a^{2}} + \frac{9}{\frac{15}{16} a^{2}} = 1 ⟹ \frac{32}{5 a^{2}} + \frac{9 \times 16}{15 a^{2}} = 1$ $\frac{32}{5 a^{2}} + \frac{48}{5 a^{2}} = 1 ⟹ \frac{80}{5 a^{2}} = 1 ⟹ \frac{16}{a^{2}} = 1 ⟹ a^{2} = 16.$
Execute: Calculate $b^{2} = \frac{15}{16} a^{2} = \frac{15}{16} \times 16 = 15$ . Thus, $a^{2} + b^{2} = 16 + 15 = 31$ .
Verify: Check: $a^{2} = 16, b^{2} = 15 ⟹ e = \sqrt{1 - 15 / 16} = 1 / 4$ . Point check: $\frac{32 / 5}{16} + \frac{9}{15} = \frac{2}{5} + \frac{3}{5} = 1$ . Both conditions satisfied exactly.

Finding $a^{2} + b^{2}$ for an Ellipse Given Eccentricity and a Point

What feels right?

Using $e^{2} = 1 - \frac{b^{2}}{a^{2}} = \frac{1}{16}$ , we find $b^{2} = \frac{15}{16} a^{2}$ ; substituting $(- 4 \sqrt{\frac{2}{5}}, 3)$ into the ellipse equation yields $a^{2} = 16$ and $b^{2} = 15$ , so $a^{2} + b^{2} = 31$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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